# The significance of special relativity

1. Jun 21, 2010

### Kholofelo

Hi

I would just like to ask a few questions bugging me.

1. Why do we regard a photon as having no-rest mass. Is the fact that a photon is 'never' at rest enough of a reason.
2. Why don't we take the photon into consideration when dealing with the gamma factor.
3. Our most sensitive equipment use light. Lets make this huge assumption that we get an object to travel at a velocity much higher than the speed of light, how then can we ever measure any speed higher than c. I hope I make no offense by this but according to me there are some parts of special relativity I am against, here is one silly argument: If we were all moles blind but not deaf. Why wouldn't special relativity revolve around the speed of sound(through air)? How would we know of light? if we stood in front of a supersonic aircraft which would someway avoid colliding with us as it approaches us, what speed would we measure? Say the craft was moving at mach 5, I will make a safe bet that our most sensitive equipment would be based on sound waves. With no doubt the sound wave which we measure would arrive only at the speed of sound. This is the quite similar to the simple scenario used to explain the basics of special relativity.
4. Following from 3, what is so special about the photon?

Thanks

2. Jun 21, 2010

### Staff: Mentor

The fact that its speed is the same for all observers is enough.
What do you mean?
This argument doesn't really hold up. Have synchronized clocks at points A and B; Send a signal from A to B, measuring its departure and arrival times; Calculate the speed of the signal. Where do you see a problem?
The fact that it moves at the apparent speed limit of the universe--the fastest possible speed.

3. Jun 21, 2010

### Kholofelo

By question 2, I meant if we were solve the gamma factor equation for a photon we have a denominator of zero given by the square-root of (1 - v^2/c^2) where v will be c since we are dealing with a photon. So what does that mean?

And sorry for my silly argument, I am still confident about about my doubts though, I am still trying to formulate a better argument to best explain my concern.

4. Jun 21, 2010

### Staff: Mentor

The gamma factor arises when comparing measurements made in frames that are in relative motion with speed v. The relative speed of such frames can never be c, thus you'd never calculate a gamma factor using v = c.

But you have a point--the fact that gamma 'blows up' if v = c tells you that you are doing something wrong. The point is that v cannot reach c.

5. Jun 21, 2010

### bcrowell

Staff Emeritus
But we don't know for sure that light travels at c. We used to think that neutrinos traveled at c, but then we found out that they had nonvanishing rest masses, so they must travel at less than c. The same could happen with the photon. All we can do is put an upper limit on the mass of the photon:

R.S. Lakes, "Experimental limits on the photon mass and cosmic magnetic vector potential", Physical Review Letters 80 (1998) 1826, http://silver.neep.wisc.edu/~lakes/mu.html

6. Jun 21, 2010

### Staff: Mentor

Experimentally, that's all we can ever do.

7. Jun 21, 2010

### Kholofelo

In a recent experiment a laser pulse was measured to be traveling at 300 times c. What justice does the gamma factor or special relativity as whole do to this scenario?

8. Jun 21, 2010

### Staff: Mentor

One must be careful in interpreting what's going on in these kinds of experiments. They generally have to do with creating a scenario in which the group velocity of a pulse is greater than c, but no one is claiming that an actual signal--which transfers information--was made to travel greater than c. There is no contradiction with relativity here.

9. Jun 21, 2010

### stevenb

Please provide references for this experiment. I agree with the above. However, without a reference, all we can do is guess.

For more details see the following.

http://en.wikipedia.org/wiki/Group_velocity

"The group velocity is often thought of as the velocity at which energy or information is conveyed along a wave. In most cases this is accurate, and the group velocity can be thought of as the signal velocity of the waveform. However, if the wave is travelling through an absorptive medium, this does not always hold. Since the 1980s, various experiments have verified that it is possible for the group velocity of laser light pulses sent through specially prepared materials to significantly exceed the speed of light in vacuum. However, superluminal communication is not possible in this case, since the signal velocity remains less than the speed of light. It is also possible to reduce the group velocity to zero, stopping the pulse, or have negative group velocity, making the pulse appear to propagate backwards. However, in all these cases, photons continue to propagate at the expected speed of light in the medium."

Last edited: Jun 21, 2010
10. Jun 21, 2010

### Al68

It's quite different. The speed of the craft would be measured at Mach 5 if sound waves were correctly used to measure its speed.

The speed of any object is limited to c as measured in an inertial reference frame, regardless of whether its measured using light or sound or just measuring tape and a clock.

11. Jun 21, 2010

### stevenb

No offense, but this is false thinking. In a sense, we ARE like blind moles. We are blind to most of the electromagnetic spectrum with the exception of a very narrow band we call visible light. We can't see or hear radio waves, microwaves, infrared rays, ultraviolet rays, gamma rays etc, yet we measure the speed with various forms of equipment. Even the speed of visible light requires special equipment to "see" its speed. Many phenomenon happen too fast or too slowly to be understood by a person even when visible observations can be made. Moreover, we use various "unseen" forms of radiation as tools to probe the universe. Radio and Xray astronomy are good examples. Also, what do observations of the cosmic microwave background radiation tell us?

Admittedly, blindness is a significant handicap to any person, including a scientist, but relying solely on what we see, hear and feel is an even bigger handicap to a scientist.

12. Jul 17, 2010

### CHUKKY

Hello i think that the issue of using a sound as a medium would still not change the implications of sr.what would happen is that the behavior of sound waves would be different or abnormal in such a situation. the issue is that since the properties of sr can be proved through light then it means that the properties which we have calculated through it would be valid irrespective of if we see. anyway we should thank God that we are not blind lest in the 21st century, the implications of sr would have been extremely difficult to discover.In science, once a phenomenon can be logically proved, the means through which it was proved is immaterial as far as that medium is independent. well the independence is not even necessary as if we were blind, we would never understand the way things operate until we are able to form some set of rules of which if traced back would revolve around the invariance of light.inn fact it is right to say that once an idea can be proved to hold valid in all situations then it is valid. Just as the same way quantum physics remains a mathematical tool though its implications are meaningless but the important issue is that quantum physics predicts the right result of all experiments it is tested with.

Last edited: Jul 17, 2010
13. Jul 18, 2010

### yuiop

Consider the equation for the momentum-energy of a particle:

$$E = \sqrt{m_o c^2 - p^2c^2}$$

where $m_o$ is the rest mass, and p is the momentum.

The momentum is usually given in relativity as:

$$p = \frac{m_o v}{\sqrt{1-v^2/c^2}}$$

For a particle with non zero rest mass the momentum becomes infinite if v=c and this tells us that it is impossible for a particle with rest mass to have v=c.

For a particle with zero rest mass the equation for energy becomes:

$$E_{zm} = \sqrt{m_o c^2 - \frac{m_o^2 v^2 c^2}{\sqrt{1-v^2/c^2}}} = \sqrt{0 c^2 - \frac{0^2 v^2 c^2}{0}} = \frac{0}{0}$$

While an infinite result means "impossible", a 0/0 result means "can not be determined using this equation, go and find another equation".

For light the momentum is given by p = hf/c where h is the Planck constant and f is the frequency. This form of the momentum can be inserted into the equation for momentum energy to give:

$$E = \sqrt{m_o c^2 + p^2c^2} = pc = hf$$

which is finite.

When used properly, the equation for momentum-energy is equally valid for particles with and without rest mass.