Relativistic Momentum - Revision history

Lheiman07 at 05:36, 30 October 2026

2026-10-30T05:36:39Z

← Older revision		Revision as of 00:36, 30 October 2026
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	:<math> \overrightarrow{p} = (1.007)(1.710^{-27})<1 10^7 , 2 * 10^7 , 3 * 10^7> = <1.7 * 10^{-20}, 3.4 * 10^{-20}, 5.1 10^{-20}> </math> kgm/s		:<math> \overrightarrow{p} = (1.007)(1.710^{-27})<1 10^7 , 2 * 10^7 , 3 * 10^7> = <1.7 * 10^{-20}, 3.4 * 10^{-20}, 5.1 10^{-20}> </math> kgm/s

	===~~Relativistic Kinetic Energy and Total Energy~~===		===Middling===

	~~In classical mechanics, the kinetic energy~~ of ~~an object~~ is ~~defined as~~		A proton has a rest mass of <math>1.67 \times 10^{-27}</math> kg and is traveling at <math>0.80c</math>, where <math>c = 3.00 \times 10^8</math> m/s.

	~~:<math>K = \frac{1}{2}mv^2</math>~~		'''(a)''' Calculate the Lorentz factor of the proton.

	~~However, this equation becomes inaccurate when an object moves at speeds comparable to~~ the ~~speed of light. Similar to momentum,~~ kinetic energy ~~must be modified to account for relativistic effects~~.		'''(b)''' Calculate the relativistic kinetic energy of the proton.

	~~The~~ relativistic ~~definition~~ of ~~total energy is~~		'''(c)''' Calculate the total relativistic energy of the proton.

	:~~<math>E = \gamma mc^2</math>~~		'''Solution:'''

	~~where <math>\gamma</math> is~~ the Lorentz factor~~, <math>m</math> is the rest mass of the particle, and <math>c</math> is the speed of light.~~		'''(a)''' First, calculate the Lorentz factor using

	An important distinction is that this equation represents the TOTAL energy of a particle, not just its kinetic energy. Even when a particle is completely at rest, it possesses energy due to its mass. This is called rest energy and is described by Einstein's famous equation:		:<math>\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}</math>

	:~~<math>E_0 = mc^2</math>~~		Substituting the given velocity:

	~~The relativistic kinetic energy is the difference between the total energy and the rest energy~~:		:<math>\gamma = \frac{1}{\sqrt{1-\frac{(0.80c)^2}{c^2}}}</math>

	:<math>K = E - ~~E_0~~</math>		:<math>= \frac{1}{\sqrt{1-0.64}}</math>

	~~Substituting the equations for total energy and rest energy gives~~		:<math>= \frac{1}{\sqrt{0.36}}</math>

	:<math>~~K =~~ \gamma ~~mc^2 - mc^2~~</math>		:<math>\boxed{\gamma = 1.667}</math>

	~~Factoring out <math>mc^2</math> gives~~ the relativistic kinetic energy ~~equation:~~		'''(b)''' Calculate the relativistic kinetic energy using

	:<math>~~\boxed{~~K = (\gamma - 1)mc^2}</math>		:<math>K = (\gamma - 1)mc^2</math>

	~~This equation demonstrates that an object's kinetic energy increases as its velocity increases and~~ the ~~Lorentz factor becomes larger.~~		Substituting the known values:

	~~'''Relationship Between Kinetic Energy and Total Energy'''~~		:<math>K = (1.667-1)(1.67 \times 10^{-27})(3.00 \times 10^8)^2</math>

	~~The total energy of a relativistic particle can be expressed as the sum of its rest energy and kinetic energy~~:		:<math>K = (0.667)(1.67 \times 10^{-27})(9.00 \times 10^{16})</math>

	:<math>E = ~~K + E_0~~</math>		:<math>\boxed{K = 1.00 \times 10^{-10} \ J}</math>

	~~:<math>E =~~ (~~\gamma - 1~~)~~mc^2 + mc^2</math>~~		'''(c)''' Calculate the total relativistic energy using

	~~Simplifying~~:		:<math>E = \gamma mc^2</math>

	:<math>\~~boxed~~{~~E =~~ \~~gamma mc~~^2}</math>		:<math>E = (1.667)(1.67 \times 10^{-27})(3.00 \times 10^8)^2</math>

	~~This relationship is important because it demonstrates that the total energy of a particle consists of both its rest energy and its kinetic energy~~.		:<math>\boxed{E = 2.50 \times 10^{-10} \ J}</math>

	~~'''What Happens at Different Velocities?'''~~		We can verify this answer by adding the kinetic energy and rest energy:

	~~There are two important limiting cases to consider:~~

	* '''When an object is at rest:''' Its velocity is zero, meaning <math>\gamma = 1</math>. Therefore, its kinetic energy is zero, but it still possesses rest energy equal to <math>mc^2</math>.

	* '''When an object approaches the speed of light:''' The Lorentz factor increases dramatically. Consequently, the kinetic energy and ~~total energy increase without bound as the velocity approaches the speed of light. This is another reason why an object with nonzero~~ rest ~~mass cannot reach the speed of light.~~

	~~'''Connection to Classical Kinetic Energy'''~~

	~~At everyday speeds, where <math>v \ll c</math>, the relativistic kinetic energy equation approaches the classical kinetic energy equation.~~

	~~In this limit, the Lorentz factor can be approximated as~~

	~~:<math>\gamma \approx 1 + \frac{v^2}{2c^2}</math>~~

	~~Substituting this approximation into the relativistic kinetic~~ energy ~~equation gives~~

	:~~<math>K = (\gamma - 1)mc^2</math>~~

	:<math>K ~~\approx \left(1~~ + ~~\frac{v^2}{2c^2} - 1\right)~~mc^2</math>		:<math>E = K + mc^2</math>

	~~Simplifying~~:		:<math>E = 1.00 \times 10^{-10} + 1.503 \times 10^{-10}</math>

	:<math>K \~~approx~~ \~~frac~~{1}{2}~~mv^2~~</math>		:<math>\boxed{E = 2.50 \times 10^{-10} \ J}</math>

	This demonstrates that ~~classical kinetic~~ energy is ~~simply an approximation~~ of relativistic kinetic energy ~~when an object's speed is much smaller than the speed of light~~.		This demonstrates that the total relativistic energy of a particle is the sum of its rest energy and relativistic kinetic energy.

	===Difficult===		===Difficult===

Lheiman07 at 05:35, 30 October 2026

2026-10-30T05:35:51Z

@@ Line 132: / Line 132: @@
 ::Plug values in
 :<math> \overrightarrow{p} = (1.007)*(1.7*10^{-27})*<1 * 10^7 , 2 * 10^7 , 3 * 10^7> = <1.7 * 10^{-20}, 3.4 * 10^{-20}, 5.1 *10^{-20}> </math> kg*m/s
 ===Difficult===

Lheiman07 at 05:34, 30 October 2026

2026-10-30T05:34:42Z

← Older revision		Revision as of 00:34, 30 October 2026
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	Another observation worth noting is that in the limit where the speed of the particle <math>v</math> approaches the speed of light <math>c</math>, the fraction <math>\frac{v^2}{c^2}</math> approaches 1 and so <math>\gamma</math> approaches infinity. This has far-reaching implications. An object of non-zero mass traveling at the speed of light would have infinite momentum according to the relativistic definition, so an infinite impulse must be applied to a particle at rest to change its speed to the speed of light. As real forces are always finite in both magnitude and duration, no particle of nonzero mass can ever achieve the speed of light. Let us analyze what happens if a constant force is applied to a particle at rest for a long time. The particle's momentum will increase linearly over time as a result of the force, as described by Newton's second law. At first, this means its speed increases linearly over time as well. However, as the particle enters the relativistic regime, <math>\gamma</math> begins to increase as well, meaning that the particle's speed doesn't have to increase as much to sustain the steady increase in momentum. Finally, when the particle's speed is nearly the speed of light, even the tiniest increases in <math>v</math> cause <math>\gamma</math> to skyrocket, meaning that the particle hardly speeds up while the particle's momentum continues to steadily increase. The speed of the particle would asymptotically approach the speed of light over time if the force continues to be applied.		Another observation worth noting is that in the limit where the speed of the particle <math>v</math> approaches the speed of light <math>c</math>, the fraction <math>\frac{v^2}{c^2}</math> approaches 1 and so <math>\gamma</math> approaches infinity. This has far-reaching implications. An object of non-zero mass traveling at the speed of light would have infinite momentum according to the relativistic definition, so an infinite impulse must be applied to a particle at rest to change its speed to the speed of light. As real forces are always finite in both magnitude and duration, no particle of nonzero mass can ever achieve the speed of light. Let us analyze what happens if a constant force is applied to a particle at rest for a long time. The particle's momentum will increase linearly over time as a result of the force, as described by Newton's second law. At first, this means its speed increases linearly over time as well. However, as the particle enters the relativistic regime, <math>\gamma</math> begins to increase as well, meaning that the particle's speed doesn't have to increase as much to sustain the steady increase in momentum. Finally, when the particle's speed is nearly the speed of light, even the tiniest increases in <math>v</math> cause <math>\gamma</math> to skyrocket, meaning that the particle hardly speeds up while the particle's momentum continues to steadily increase. The speed of the particle would asymptotically approach the speed of light over time if the force continues to be applied.

			===Relativistic Kinetic Energy and Total Energy===

			In classical mechanics, the kinetic energy of an object is defined as

			:<math>K = \frac{1}{2}mv^2</math>

			However, this equation becomes inaccurate when an object moves at speeds comparable to the speed of light. Similar to momentum, kinetic energy must be modified to account for relativistic effects.

			The relativistic definition of total energy is

			:<math>E = \gamma mc^2</math>

			where <math>\gamma</math> is the Lorentz factor, <math>m</math> is the rest mass of the particle, and <math>c</math> is the speed of light.

			An important distinction is that this equation represents the TOTAL energy of a particle, not just its kinetic energy. Even when a particle is completely at rest, it possesses energy due to its mass. This is called rest energy and is described by Einstein's famous equation:

			:<math>E_0 = mc^2</math>

			The relativistic kinetic energy is the difference between the total energy and the rest energy:

			:<math>K = E - E_0</math>

			Substituting the equations for total energy and rest energy gives

			:<math>K = \gamma mc^2 - mc^2</math>

			Factoring out <math>mc^2</math> gives the relativistic kinetic energy equation:

			:<math>\boxed{K = (\gamma - 1)mc^2}</math>

			This equation demonstrates that an object's kinetic energy increases as its velocity increases and the Lorentz factor becomes larger.

			'''Relationship Between Kinetic Energy and Total Energy'''

			The total energy of a relativistic particle can be expressed as the sum of its rest energy and kinetic energy:

			:<math>E = K + E_0</math>

			:<math>E = (\gamma - 1)mc^2 + mc^2</math>

			Simplifying:

			:<math>\boxed{E = \gamma mc^2}</math>

			This relationship is important because it demonstrates that the total energy of a particle consists of both its rest energy and its kinetic energy.

			'''What Happens at Different Velocities?'''

			There are two important limiting cases to consider:

			* '''When an object is at rest:''' Its velocity is zero, meaning <math>\gamma = 1</math>. Therefore, its kinetic energy is zero, but it still possesses rest energy equal to <math>mc^2</math>.

			* '''When an object approaches the speed of light:''' The Lorentz factor increases dramatically. Consequently, the kinetic energy and total energy increase without bound as the velocity approaches the speed of light. This is another reason why an object with nonzero rest mass cannot reach the speed of light.

			'''Connection to Classical Kinetic Energy'''

			At everyday speeds, where <math>v \ll c</math>, the relativistic kinetic energy equation approaches the classical kinetic energy equation.

			In this limit, the Lorentz factor can be approximated as

			:<math>\gamma \approx 1 + \frac{v^2}{2c^2}</math>

			Substituting this approximation into the relativistic kinetic energy equation gives

			:<math>K = (\gamma - 1)mc^2</math>

			:<math>K \approx \left(1 + \frac{v^2}{2c^2} - 1\right)mc^2</math>

			Simplifying:

			:<math>K \approx \frac{1}{2}mv^2</math>

			This demonstrates that classical kinetic energy is simply an approximation of relativistic kinetic energy when an object's speed is much smaller than the speed of light.

	==Examples==		==Examples==

YorickAndeweg: /* Difficult */

2019-08-03T19:11:38Z

Difficult

← Older revision		Revision as of 14:11, 3 August 2019
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	<math> \vec{p} = 3.203 * 1.7 * 10^{-27} * (.95 * 3 * 10^8) = 1.55 * 10^{-18} </math> kg*m/s		<math> \vec{p} = 3.203 * 1.7 * 10^{-27} * (.95 * 3 * 10^8) = 1.55 * 10^{-18} </math> kg*m/s

	Now let us find the momentum at the end of the impulse using the impulse-momentum theorem ~~(see [[impulse and momentum~~]]).		Now let us find the momentum at the end of the impulse using the [[Impulse and Momentum\|impulse-momentum theorem]].

	<math> \vec{p}_f = \vec{p}_i + \vec{J} </math>		<math> \vec{p}_f = \vec{p}_i + \vec{J} </math>

YorickAndeweg: /* The Main Idea */

2019-08-03T19:11:10Z

The Main Idea

← Older revision		Revision as of 14:11, 3 August 2019
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	==The Main Idea==		==The Main Idea==

	The [[Linear Momentum]] of an object is traditionally defined as <math>\vec{p} = m \vec{v}</math>. Momentum is a useful quantity because many other concepts relate to it. For example, forces cause the momentum of a system to change according to [[Newton's Second Law: the Momentum Principle]], and in certain situations, [[Conservation of Momentum]]\|the momentum of a system is conserved]]. However, for systems containing objects moving at speeds comparable to the speed of light, both Newton's second law and conservation of momentum appear to be violated. As it turns out, if the definition of the momentum of a particle is adjusted to accommodate high-speed objects, these relations hold true after all. The adjusted definition of momentum is called the relativistic definition and defines momentum as follows:		The [[Linear Momentum]] of an object is traditionally defined as <math>\vec{p} = m \vec{v}</math>. Momentum is a useful quantity because many other concepts relate to it. For example, forces cause the momentum of a system to change according to [[Newton's Second Law: the Momentum Principle]], and in certain situations, [[Conservation of Momentum\|the momentum of a system is conserved]]. However, for systems containing objects moving at speeds comparable to the speed of light, both Newton's second law and conservation of momentum appear to be violated. As it turns out, if the definition of the momentum of a particle is adjusted to accommodate high-speed objects, these relations hold true after all. The adjusted definition of momentum is called the relativistic definition and defines momentum as follows:

	<math> \vec{p} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} * m \vec{v} </math>		<math> \vec{p} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} * m \vec{v} </math>

YorickAndeweg: /* The Main Idea */

2019-08-03T19:10:48Z

The Main Idea

← Older revision		Revision as of 14:10, 3 August 2019
Line 3:		Line 3:
	==The Main Idea==		==The Main Idea==

	The [[Linear Momentum]] of an object is traditionally defined as <math>\vec{p} = m \vec{v}</math>. Momentum is a useful quantity because many other concepts relate to it. For example, forces cause the momentum of a system to change according to [[Newton's Second Law: the Momentum Principle]], and in certain situations, the momentum of a system is conserved ~~(see [[Conservation of Momentum~~]]). However, for systems containing objects moving at speeds comparable to the speed of light, both Newton's second law and conservation of momentum appear to be violated. As it turns out, if the definition of the momentum of a particle is adjusted to accommodate high-speed objects, these relations hold true after all. The adjusted definition of momentum is called the relativistic definition and defines momentum as follows:		The [[Linear Momentum]] of an object is traditionally defined as <math>\vec{p} = m \vec{v}</math>. Momentum is a useful quantity because many other concepts relate to it. For example, forces cause the momentum of a system to change according to [[Newton's Second Law: the Momentum Principle]], and in certain situations, [[Conservation of Momentum]]\|the momentum of a system is conserved]]. However, for systems containing objects moving at speeds comparable to the speed of light, both Newton's second law and conservation of momentum appear to be violated. As it turns out, if the definition of the momentum of a particle is adjusted to accommodate high-speed objects, these relations hold true after all. The adjusted definition of momentum is called the relativistic definition and defines momentum as follows:

	<math> \vec{p} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} * m \vec{v} </math>		<math> \vec{p} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} * m \vec{v} </math>

YorickAndeweg: /* The Main Idea */

2019-08-03T19:08:51Z

The Main Idea

← Older revision		Revision as of 14:08, 3 August 2019
Line 3:		Line 3:
	==The Main Idea==		==The Main Idea==

	The [[Linear ~~momentum~~]] of an object is traditionally defined as <math>\vec{p} = m \vec{v}</math>. Momentum is a useful quantity because many other concepts relate to it. For example, forces cause the momentum of a system to change according to [[Newton's Second Law: the Momentum Principle]], and in certain situations, the momentum of a system is conserved (see [[Conservation of Momentum]]). However, for systems containing objects moving at speeds comparable to the speed of light, both Newton's second law and conservation of momentum appear to be violated. As it turns out, if the definition of the momentum of a particle is adjusted to accommodate high-speed objects, these relations hold true after all. The adjusted definition of momentum is called the relativistic definition and defines momentum as follows:		The [[Linear Momentum]] of an object is traditionally defined as <math>\vec{p} = m \vec{v}</math>. Momentum is a useful quantity because many other concepts relate to it. For example, forces cause the momentum of a system to change according to [[Newton's Second Law: the Momentum Principle]], and in certain situations, the momentum of a system is conserved (see [[Conservation of Momentum]]). However, for systems containing objects moving at speeds comparable to the speed of light, both Newton's second law and conservation of momentum appear to be violated. As it turns out, if the definition of the momentum of a particle is adjusted to accommodate high-speed objects, these relations hold true after all. The adjusted definition of momentum is called the relativistic definition and defines momentum as follows:

	<math> \vec{p} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} * m \vec{v} </math>		<math> \vec{p} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} * m \vec{v} </math>

YorickAndeweg: /* Examples */

2019-07-07T18:27:42Z

Examples

← Older revision		Revision as of 13:27, 7 July 2019
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	Suppose that a proton (mass = <math> 1.7 * 10^{-27} </math> kg) is moving with a speed of <math> .97c </math> <br>		Suppose that a proton (mass = <math> 1.7 * 10^{-27} </math> kg) is moving with a speed of <math> .97c </math> <br>
	<br> What is the magnitude of the momentum of the proton?		<br> What is the magnitude of the momentum of the proton?

			Solution:

	:<math> \frac{v}{c} = \frac{.97c}{c}</math> <math> = .97 </math>		:<math> \frac{v}{c} = \frac{.97c}{c}</math> <math> = .97 </math>
	:<math> \gamma = \frac{1}{\sqrt{1-(.97)^2}} = 4.1135 </math>		:<math> \gamma = \frac{1}{\sqrt{1-(.97)^2}} = 4.1135 </math>
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	Suppose that a proton (mass = <math> 1.7 * 10^{-27} </math> kg) is moving with a velocity <math> <1 * 10^7 , 2 * 10^7 , 3 * 10^7> </math> m/s. <br>		Suppose that a proton (mass = <math> 1.7 * 10^{-27} </math> kg) is moving with a velocity <math> <1 * 10^7 , 2 * 10^7 , 3 * 10^7> </math> m/s. <br>
	<br> What is the momentum of the proton?		<br> What is the momentum of the proton?

			Solution:

	:<math> \left\vert \overrightarrow{v} \right\vert = \sqrt{(110^7)^2 + (210^7)^2 +(310^7)^2} </math> m/s <math> = 3.7 10^7 </math> m/s.		:<math> \left\vert \overrightarrow{v} \right\vert = \sqrt{(110^7)^2 + (210^7)^2 +(310^7)^2} </math> m/s <math> = 3.7 10^7 </math> m/s.
	:<math> \frac{\left\vert \overrightarrow{v} \right\vert}{c} = \frac{3.7 * 10^7 m/s}{3 * 10^8 m/s}</math> <math> = .12 </math>		:<math> \frac{\left\vert \overrightarrow{v} \right\vert}{c} = \frac{3.7 * 10^7 m/s}{3 * 10^8 m/s}</math> <math> = .12 </math>
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	Suppose that a proton (mass = <math> 1.7 * 10^{-27} </math> kg) is moving at a speed of <math>.95c</math>. A force of magnitude <math>2 * 10^{-17}</math>N acts on it for .5s in the direction of the proton's velocity, causing it to speed up. What is its new speed at the end of the .5s time interval? Use <math>3*10^8</math> for <math>c</math>.		Suppose that a proton (mass = <math> 1.7 * 10^{-27} </math> kg) is moving at a speed of <math>.95c</math>. A force of magnitude <math>2 * 10^{-17}</math>N acts on it for .5s in the direction of the proton's velocity, causing it to speed up. What is its new speed at the end of the .5s time interval? Use <math>3*10^8</math> for <math>c</math>.

			Solution:

	Let us first find the initial momentum of the particle:		Let us first find the initial momentum of the particle:

YorickAndeweg: /* References */

2019-06-07T02:26:27Z

References

← Older revision		Revision as of 21:26, 6 June 2019
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	==References==		==References==
	*https://opentextbc.ca/physicstestbook2/chapter/relativistic-momentum/		*https://opentextbc.ca/physicstestbook2/chapter/relativistic-momentum/

			[[Category:Momentum]]

YorickAndeweg: /* Connectedness */

2019-06-04T18:38:45Z

Connectedness

← Older revision		Revision as of 13:38, 4 June 2019
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	==Connectedness==		==Connectedness==

	Relativistic momentum has applications in quantum physics, particle physics, and astrophysics. In these branches of physics, particles routinely travel at relativistic speeds, and often, their momenta are directly measurable by the particles' interactions with matter. In industry, relativistic momentum is mostly used by ~~the~~ space flight ~~industry~~.		Relativistic momentum has applications in quantum physics, particle physics, and astrophysics. In these branches of physics, particles routinely travel at relativistic speeds, and often, their momenta are directly measurable by the particles' interactions with matter. In industry, relativistic momentum is mostly used by space flight institutions.

	==History==		==History==