# Light without mass

1. Apr 30, 2006

### Mt. Nixion

Hi, I read this from a website, which says,

INFINITE ACCELERATION: Acceleration (according to General Relativity) has the effect of dilating time. A photon undergoing infinite acceleration/deceleration would 'experience' no elapsed time on a journey of any spatial length.At infinite acceleration one can travel anywhere in no time, just as one can be transported anywhere while experiencing the infinite acceleration of an Ancient Future dance performance.

In classical mechanics, zero mass would result in infinite acceleration. Now light has no mass (meanng it has no inertia), yet it has a constant speed.

Therefore, my question is how can light travel at a constant speed and have no infinite acceleration when it does not have any mass or inertia? How can light have no mass and yet behave as if it did have mass?

2. Apr 30, 2006

### DaveC426913

Ok, the website spews forth nonsense - ignore it. But there are grains of truth in what it says.

A photon moves at the speed of light and has 0 acceleration.

A photon does not experience time. Every point the universe it visits it does so simultaneously - form "its" perspective (inasmuch as a photon *has* a "persepctive" to experience anything at all).

3. Apr 30, 2006

### Hootenanny

Staff Emeritus
You cannot apply the concepts of classical mechanics to a photon, Classical mechanics is superceeded by relativity and quantum mechanics.

~H

4. Apr 30, 2006

### Mt. Nixion

You are truly right, my friend. Can you explain why light has a constant speed, please?

5. Apr 30, 2006

### pmb_phy

Applying a pulse of localized radiation is the same thing as far as SR is concerned. Replace "photon" with "pulse of light" and there will be no problems addressing the question posed.
Light only accelerates in a gravitational field. While light has zero rest mass it has a non-zero relativistic mass.

Pete

6. Apr 30, 2006

### rbj

i think there are theoretical reasons for why the speed of E&M propagation (of which light is) should be the same for all inertial reference frames. it really just comes from Maxwell's Eqs. and the knowledge that there is no aether medium that E&M is propagated in.

for sound, this is wave propagation within air by minute compressions and rarefractions of the gas and we can tell if the medium, air, is moving past us or not. and if it is moving past us, we can observe a different speed of sound in the direction of the wind compared to the opposite direction.

but how do we tell the difference between a moving vacuum and a stationary vacuum? if we can't, if there really is no difference between a moving vacuum and a stationary vacuum, that such a concept is really meaningless, then whether the light that you are measuring originated from a flashlight mounted on a rocket moving past you at $c/2$ or from a stationary flashlight, how does that change the fact that a changing E field is causing a changing B field which is causing a changing E field which is causing a changing B field which is causing a changing E field, etc.? that propagation of an E field and B field disturbance, which, from solving Maxwell's Equations, has velocity $$c = 1 / \sqrt{ \epsilon_0 \mu_0 }$$? two observers moving past each other at high speed in a vacuum, only one of them is holding the flashlight, are looking at the same little beam of light in a vacuum, the same propagation of E and B fields. once the wave has left the flashlight, it doesn't know that it "had a boost" from the speed of the flashlight, it is just propagating through a vacuum. both observers have equal claim to Maxwell's Equations. both observers will see it propagate at a speed of $$c = 1 / \sqrt{ \epsilon_0 \mu_0 }$$.

how is it different? whether you are holding the flashlight or moving past it at high velocity, Maxwell's Eqs. say the same thing regarding the nature of E&M in the vacuum.

so then, there is no reason to expect a different observed speed of light for the different observers (the one on the rocket and the other who is "stationary").