Does a photon in vacuum have a rest frame?

Bernard Boyer
2022-07-23
Answered

Does a photon in vacuum have a rest frame?

You can still ask an expert for help

akademiks1989rz

Answered 2022-07-24
Author has **16** answers

Short answer: no.

Explanation:

Many introductory text books talk about "rest mass" and "relativistic mass" and say that the "rest mass" is the mass measured in the particles rest frame.

That's not wrong, you can do physics in that point of view, but that is not how people talk about and define mass anymore.

In the modern view each particle has one and only one mass defined by the square of it's energy--momentum four vector (which being a Lorentz invariant you can calculate in any inertial frame):

${m}^{2}\equiv {p}^{2}=(E,\overrightarrow{p}{)}^{2}={E}^{2}-{\overrightarrow{p}}^{2}$

For a photon this value is zero. In any frame, and that allows people to reasonably say that the photon has zero mass without needing to define a rest frame for it.

Explanation:

Many introductory text books talk about "rest mass" and "relativistic mass" and say that the "rest mass" is the mass measured in the particles rest frame.

That's not wrong, you can do physics in that point of view, but that is not how people talk about and define mass anymore.

In the modern view each particle has one and only one mass defined by the square of it's energy--momentum four vector (which being a Lorentz invariant you can calculate in any inertial frame):

${m}^{2}\equiv {p}^{2}=(E,\overrightarrow{p}{)}^{2}={E}^{2}-{\overrightarrow{p}}^{2}$

For a photon this value is zero. In any frame, and that allows people to reasonably say that the photon has zero mass without needing to define a rest frame for it.

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 2022-05-20

It would be nice to have a cute method that uses Lorentz transformations of basis vectors by exponential transformation using gamma matrices. To avoid confusion, let's assume $-+++$ signature. Given ${\gamma}_{\mu}$ as gamma matrices that satisfy $\mathrm{t}\mathrm{r}({\gamma}_{\mu}{\gamma}_{\nu})=4{\eta}_{\mu \nu}$, then we have $\mathrm{t}\mathrm{r}({\gamma}_{\mu}^{\prime}{\gamma}_{\nu}^{\prime})=4{\eta}_{\mu \nu}$ if we put:${\gamma}_{\mu}^{\prime}=\mathrm{exp}(-A){\gamma}_{\mu}\mathrm{exp}(+A)$where $A$ is any matrix.

Boosts and rotations use $A$ as a bivector. For example, with $\alpha $ a real number, $A=\alpha {\gamma}_{0}{\gamma}_{3}$ boosts in the $z$ direction while $A=\alpha {\gamma}_{1}{\gamma}_{2}$ gives a rotation around the $z$ axis.

Solving for the value of α that gives a boost with velocity $\beta =v/c$ appears to be straightforward. But how to do the rotations?

And by the way, what happens when you generalize $A$ to be something other than bivectors?

Boosts and rotations use $A$ as a bivector. For example, with $\alpha $ a real number, $A=\alpha {\gamma}_{0}{\gamma}_{3}$ boosts in the $z$ direction while $A=\alpha {\gamma}_{1}{\gamma}_{2}$ gives a rotation around the $z$ axis.

Solving for the value of α that gives a boost with velocity $\beta =v/c$ appears to be straightforward. But how to do the rotations?

And by the way, what happens when you generalize $A$ to be something other than bivectors?

asked 2022-05-09

For a car that is accelerating linearly, the non-inertial frame of reference is the driver in the car where from his reference frame, the car is stationary. It is so called stationary because the non-inertial frame of reference has the same acceleration as the car. Is like the car's acceleration "transform" the driver frame of reference into a non-inertial. That's why in the non-inertial frame of reference, there is no force acting on the car.

But when the car is driving in circles at a constant speed, in the non-inertial frame of reference there is a force acting on the car, which is the centripetal force. Why isn't this frame of reference like the above, not having the acceleration found in their each respective inertial reference frame? Why can't we have a non-inertial reference frame(due to rotation) whereby there is no centripetal force, subsequently eradicating the need for a centrifugal force?

But when the car is driving in circles at a constant speed, in the non-inertial frame of reference there is a force acting on the car, which is the centripetal force. Why isn't this frame of reference like the above, not having the acceleration found in their each respective inertial reference frame? Why can't we have a non-inertial reference frame(due to rotation) whereby there is no centripetal force, subsequently eradicating the need for a centrifugal force?

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 $\frac{1}{2}m{v}^{2}$.

asked 2022-07-20

Does the temperature of a body depend on the frame of reference?

asked 2022-05-15

What is a homogeneous and isotropic frame of reference?

I have heard that inertial frames of reference in the context of special relativity are both isotropic and homogeneous. I know what isotropic and homogeneous mean in a general context, but what do they mean when relating to a frame of reference?

I have heard that inertial frames of reference in the context of special relativity are both isotropic and homogeneous. I know what isotropic and homogeneous mean in a general context, but what do they mean when relating to a frame of reference?

asked 2022-05-13

A rotating frame of reference (since you rotate your BEC to generate these), where the energy is given by:

$\stackrel{~}{E}[\mathrm{\Psi}]=E[\mathrm{\Psi}]-L[\mathrm{\Psi}]\cdot \mathrm{\Omega},$

Where $\mathrm{\Omega}$ is the rotational velocity which you apply to the BEC.

Now the argument that the term $L[\mathrm{\Psi}]\cdot \mathrm{\Omega}$ should be substracted comes from the fact that in a rotating frame your system loses that fraction of rotational energy. Now I was wondering if anyone knew where the form of $L[\mathrm{\Psi}]\cdot \mathrm{\Omega}$ came from?

I know that in a rotating frame of reference you have that $\overrightarrow{v}={\overrightarrow{v}}_{r}+\overrightarrow{\mathrm{\Omega}}\times \overrightarrow{r}$ . If you fill this in into the kinetic energy and use some basic definitions, you get that the extra effect of rotation is given by:

$\frac{1}{2}I{\mathrm{\Omega}}^{2}=\frac{1}{2}J\mathrm{\Omega}.$

This yields half of the value that is used in the book, is there something that I'm missing or not seeing right ?

$\stackrel{~}{E}[\mathrm{\Psi}]=E[\mathrm{\Psi}]-L[\mathrm{\Psi}]\cdot \mathrm{\Omega},$

Where $\mathrm{\Omega}$ is the rotational velocity which you apply to the BEC.

Now the argument that the term $L[\mathrm{\Psi}]\cdot \mathrm{\Omega}$ should be substracted comes from the fact that in a rotating frame your system loses that fraction of rotational energy. Now I was wondering if anyone knew where the form of $L[\mathrm{\Psi}]\cdot \mathrm{\Omega}$ came from?

I know that in a rotating frame of reference you have that $\overrightarrow{v}={\overrightarrow{v}}_{r}+\overrightarrow{\mathrm{\Omega}}\times \overrightarrow{r}$ . If you fill this in into the kinetic energy and use some basic definitions, you get that the extra effect of rotation is given by:

$\frac{1}{2}I{\mathrm{\Omega}}^{2}=\frac{1}{2}J\mathrm{\Omega}.$

This yields half of the value that is used in the book, is there something that I'm missing or not seeing right ?