Accelerating particles to speeds infinitesimally close to the speed of light?

lifretatox8n
2022-04-07
Answered

Accelerating particles to speeds infinitesimally close to the speed of light?

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candydulce168nlid

Answered 2022-04-08
Author has **14** answers

By special relativity, the energy needed to accelerate a particle (with mass) grow super-quadratically when the speed is close to c, and is $\infty $ when it is $c$.

$E=\gamma m{c}^{2}=\frac{m{c}^{2}}{\sqrt{1-(\text{\u201cpercent of speed of light\u201d}{)}^{2}}}$

Since you can't supply infinite energy to the particle, it is not possible to get to $100\mathrm{\%}\text{}c$.

$E=\gamma m{c}^{2}=\frac{m{c}^{2}}{\sqrt{1-(\text{\u201cpercent of speed of light\u201d}{)}^{2}}}$

Since you can't supply infinite energy to the particle, it is not possible to get to $100\mathrm{\%}\text{}c$.

asked 2022-05-18

I am asked to show that the matrix

$S=1-\frac{i}{4}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}$

represents the infinitesimal Lorentz transformation

${\mathrm{\Lambda}}^{\mu}{}_{\nu}={\delta}^{\mu}{}_{\nu}+{\epsilon}^{\mu}{}_{\nu},$

in the sense that

${S}^{-1}{\gamma}^{\mu}S={\mathrm{\Lambda}}^{\mu}{}_{\nu}{\gamma}^{\nu}.$

I have already proven that ${S}^{-1}={\gamma}^{0}{S}^{\u2020}{\gamma}^{0}$, so I can begin with the left-hand side of this third equation:

$\begin{array}{rl}{S}^{-1}{\gamma}^{\mu}S& ={\gamma}^{0}{S}^{\u2020}{\gamma}^{0}{\gamma}^{\mu}S\\ & ={\gamma}^{0}(1+\frac{i}{4}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}){\gamma}^{0}{\gamma}^{\mu}(1-\frac{i}{4}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu})\\ & ={\gamma}^{0}{\gamma}^{0}{\gamma}^{\mu}-{\gamma}^{0}{\gamma}^{0}{\gamma}^{\mu}\frac{i}{4}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}+{\gamma}^{0}\frac{i}{4}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}{\gamma}^{0}{\gamma}^{\mu}-{\gamma}^{0}\frac{i}{4}{\sigma}_{\rho \tau}{\epsilon}^{\rho \tau}{\gamma}^{0}{\gamma}^{\mu}\frac{i}{4}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}\\ & ={\gamma}^{\mu}+\frac{1}{16}{\gamma}^{0}{\sigma}_{\rho \tau}{\epsilon}^{\rho \tau}{\gamma}^{0}{\gamma}^{\mu}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}\\ & ={\gamma}^{\mu}+\frac{1}{16}\left({\sigma}_{\rho \tau}{\epsilon}^{\rho \tau}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}\right){\gamma}^{\mu}.\end{array}$

Remember, we want this to be equal to

${\mathrm{\Lambda}}^{\mu}{}_{\nu}{\gamma}^{\nu}=({\delta}^{\mu}{}_{\nu}+{\epsilon}^{\mu}{}_{\nu}){\gamma}^{\nu}={\gamma}^{\mu}+{\epsilon}^{\mu}{}_{\nu}{\gamma}^{\nu}.$

The first term is there already, but I have no idea how that second term is going to work out to ${\epsilon}^{\mu}{}_{\nu}{\gamma}^{\nu}$. Can anyone give me a hint, or tell me what I’ve done wrong?

$S=1-\frac{i}{4}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}$

represents the infinitesimal Lorentz transformation

${\mathrm{\Lambda}}^{\mu}{}_{\nu}={\delta}^{\mu}{}_{\nu}+{\epsilon}^{\mu}{}_{\nu},$

in the sense that

${S}^{-1}{\gamma}^{\mu}S={\mathrm{\Lambda}}^{\mu}{}_{\nu}{\gamma}^{\nu}.$

I have already proven that ${S}^{-1}={\gamma}^{0}{S}^{\u2020}{\gamma}^{0}$, so I can begin with the left-hand side of this third equation:

$\begin{array}{rl}{S}^{-1}{\gamma}^{\mu}S& ={\gamma}^{0}{S}^{\u2020}{\gamma}^{0}{\gamma}^{\mu}S\\ & ={\gamma}^{0}(1+\frac{i}{4}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}){\gamma}^{0}{\gamma}^{\mu}(1-\frac{i}{4}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu})\\ & ={\gamma}^{0}{\gamma}^{0}{\gamma}^{\mu}-{\gamma}^{0}{\gamma}^{0}{\gamma}^{\mu}\frac{i}{4}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}+{\gamma}^{0}\frac{i}{4}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}{\gamma}^{0}{\gamma}^{\mu}-{\gamma}^{0}\frac{i}{4}{\sigma}_{\rho \tau}{\epsilon}^{\rho \tau}{\gamma}^{0}{\gamma}^{\mu}\frac{i}{4}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}\\ & ={\gamma}^{\mu}+\frac{1}{16}{\gamma}^{0}{\sigma}_{\rho \tau}{\epsilon}^{\rho \tau}{\gamma}^{0}{\gamma}^{\mu}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}\\ & ={\gamma}^{\mu}+\frac{1}{16}\left({\sigma}_{\rho \tau}{\epsilon}^{\rho \tau}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}\right){\gamma}^{\mu}.\end{array}$

Remember, we want this to be equal to

${\mathrm{\Lambda}}^{\mu}{}_{\nu}{\gamma}^{\nu}=({\delta}^{\mu}{}_{\nu}+{\epsilon}^{\mu}{}_{\nu}){\gamma}^{\nu}={\gamma}^{\mu}+{\epsilon}^{\mu}{}_{\nu}{\gamma}^{\nu}.$

The first term is there already, but I have no idea how that second term is going to work out to ${\epsilon}^{\mu}{}_{\nu}{\gamma}^{\nu}$. Can anyone give me a hint, or tell me what I’ve done wrong?

asked 2022-08-12

The argument is often given that the early attempts of constructing a relativistic theory of quantum mechanics must not have gotten everything right because they led to the necessity of negative energy states. What's so wrong with that? Why can't we have negative energy states?

asked 2022-04-06

If you have 2 flashlights, one facing North and one facing South, how fast are the photons (or lightbeams) from both flashlights moving away from one another?

Just adding speeds would yield 2C, but that's not possible as far as I know.

The reference frame here would be the place where the flashlights are and/or .The beams relative to one another.

Just adding speeds would yield 2C, but that's not possible as far as I know.

The reference frame here would be the place where the flashlights are and/or .The beams relative to one another.

asked 2022-08-17

Two person, $A$ and $B$, each holding one end of a long solid rod.

Now person $A$ pushes the rod on one end.

Question: Is it correct that the information that the rod has been pushed will travel to the other end at the speed of light whereas the actual 'push' will travel at the speed of sound in the rod?

Now person $A$ pushes the rod on one end.

Question: Is it correct that the information that the rod has been pushed will travel to the other end at the speed of light whereas the actual 'push' will travel at the speed of sound in the rod?

asked 2022-07-22

When an electron emits a photon from changing energy levels, the frequency of the photon depends on the difference between the energy levels.

But if someone is moving with respect to the atom, the frequency will be apparently red shifted or blue shifted.

Does this mean that the energy levels of the orbits of the atom look to have different values if you are moving with respect to them? But, the apparent energy difference can be different depending on whether the photon is emitted towards you or away from you.

What's going on? Aren't energy levels of orbits supposed to be a fixed value given the kind of atom?

But if someone is moving with respect to the atom, the frequency will be apparently red shifted or blue shifted.

Does this mean that the energy levels of the orbits of the atom look to have different values if you are moving with respect to them? But, the apparent energy difference can be different depending on whether the photon is emitted towards you or away from you.

What's going on? Aren't energy levels of orbits supposed to be a fixed value given the kind of atom?

asked 2022-08-24

Is there any way in which a bound state could consist only of massless particles? If yes, would this "atom" of massless particles travel on a light-like trajectory, or would the interaction energy cause it to travel on a time-like trajectory?

asked 2022-07-14

If the velocity is a relative quantity, will it make inconsistent equations when applying it to the conservation of energy equations?

For example:

In the train moving at $V$ relative to ground, there is an object moving at $v$ relative to the frame in the same direction the frame moves. Observer on the ground calculates the object kinetic energy as $\frac{1}{2}m(v+V{)}^{2}$. However, another observer on the frame calculates the energy as $\frac{1}{2}m{v}^{2}$.

For example:

In the train moving at $V$ relative to ground, there is an object moving at $v$ relative to the frame in the same direction the frame moves. Observer on the ground calculates the object kinetic energy as $\frac{1}{2}m(v+V{)}^{2}$. However, another observer on the frame calculates the energy as $\frac{1}{2}m{v}^{2}$.