# Question About Speed of Light and Energy

1. Apr 25, 2007

### Arepo

Alright, so according to General Relativity, nothing can go faster than the speed of light in a vacuum. And is this because, as an object goes faster, it gains more mass and you need more energy to accel even more, but what causes this? Is it because energy has mass?
I realize there are probably a lot of things I am oblivious to, but any kind of answer would be appreciated.

2. Apr 25, 2007

### mezarashi

You are probably talking about the special theory of relativity. The time dilation, length contraction, etc occurs because firstly... it's true, it really happens. But from the theoretical frame work, if you rederive all the motion equations (the stuff Newton did) based on Einstein's 2 new postulates, these strange phenomena occur.

1. Rules of physics are fair in all reference frames
2. The speed of light is invariant, meaning light always travels at the speed of light regardless of how you are moving.

Now in order for the 2nd postulate to be true, it's sometimes necessary for time to dilate and lengths to contract.

3. Apr 25, 2007

### Arepo

Ah, alright thanks, I don't know much about the topic, but it is really interesting.
But one question it doesn't seem you have answered is, does that mean energy has mass?

4. Apr 25, 2007

### mezarashi

In many ways you can say that energy and mass are "interchangeable". You start to move from conservation of energy to conservation of energy-mass instead, but the story is much more complicated than that.

For example, photons (particles of light) have momentum. But how can particles of energy with no mass have momentum? But how can something with mass move at the speed of light then? Wait, but this IS light itself! The story continues...

In any case, I'm sure you've at least seen E=mc^2 before ^^.

5. Apr 25, 2007

### bernhard.rothenstein

energy mass

Start with m=m(0/sqrt(1-VV/cc) an equation which accounts for a well known experimental fact (Bucherer, Kaufmann). Multiply its both sides with cc in order to obtain E=E(0)/sqrt(1-VV/cc). Is there more to say? Use those equations when you consider interactions between particles from two inertial reference frames in order to obtain correct results.
soft words and hard arguments

6. Apr 25, 2007

### Arepo

Can you explain that a little more indepth?

7. Apr 25, 2007

### Xezlec

As I understand it, mass and energy are the same "stuff". Energy gravitates just like mass, and energy has inertia just like mass. It can be shown mathematically (I've done so when bored) that the increasing "difficulty" of accelerating an object at higher and higher velocities is what would be expected once you take into account the extra inertia of all the kinetic energy of the object. I'm sure that's not the "official" explanation, but the algebra does work out exactly right. So I'd say "yes" to your question there. :-)

EDIT: let me make one note here -- the way you phrased that, "energy has mass," isn't really the right way to say it. I know what you mean, and I was answering "yes" to what I think you intended to ask. However, "mass" is used to mean the stuff that is the same no matter how fast it's going. "Energy" is the other stuff, like kinetic energy. Even though both mass and energy are the same kind of "stuff" that behaves the same way, we use the two terms in different contexts.

Last edited: Apr 25, 2007
8. Apr 26, 2007

### pervect

Staff Emeritus
One often hears this explanation in the popular literature, and while it is mostly correct, it rather obscures some important points.

This explanation relies on dynamics (forces & mass) to explain kinematics (why you can't go faster than light).

There is a more direct explanation. Velocities in SR simply don't add linearly. This happens because of relativistic effects (length contraction, time dilation). So if you accelerate at 10 meters per second^2, you do not always add 10 meters to your velocity relative to your starting point in one second.

In fact, as you approach the speed of light, you add less and less, and it turns out that you never exceed the speed of light.

9. Apr 26, 2007

### Jonathan Scott

In classical physics, we think of time and distance as being independent. In special relativity, we find that the time between two points on the path of a travelling object as measured in the usual way effectively also includes the distance between the events, in that mathematically time is like the hypotenuse (long side) of a right-angled triangle where the other two sides are equal to the distance travelled and the amount the travelling object has aged. This means that however fast an object travels, all it can do is approach the case where the distance travelled is equal to the time and the travelling object does not get any older, which is the speed of light.

Note however that there is no limit on the displacement in space of an object divided by the amount that it has aged, which is similar to the classical velocity for small speeds, but diverges at relativistic speeds. This ratio is known as the "proper velocity" of the object, and in some ways behaves rather more as one would expect the classical velocity to behave, in that for example it can be multiplied by the rest mass to give the correct linear momentum.

10. Apr 26, 2007

### MeJennifer

The amount the traveling object has aged is actually represented by the hypotenuse not by one of the sides.

In relativity there is no such thing as absolute space. The term displacement of space is meaningless in relativity.

There are an infinite number of possible timelike paths of different lengths between two given events. The length of each of those paths represents a particular time between those events. In other words times vary for different paths between two events.

Last edited: Apr 26, 2007
11. Apr 26, 2007

### Jonathan Scott

No, you've misunderstood what I was saying; the proper time is of course the magnitude of a space-time displacement, but that uses a different sign convention.

If the amount of aging (proper time) is s and we are using units such that c=1 then both of the following are true:

ds^2 = dt^2 - dx^2

dt^2 = ds^2 + dx^2

The first form is of course the conventional magnitude of a space-time separation. I was using the second merely to illustrate that the triangle inequality that dt for two events on the same path is at least equal to dx.

For a more intuitive picture of space-time separations, consider some pieces of string which are joined together. The spatial displacements x1, x2 between the ends of each piece add as vectors to give the overall displacement x1+x2 between the outer ends of the joined pieces, and the lengths t1, t2 add to give the total length t1+t2. If you hold the end points of each piece fixed, then the middle of that piece can be moved to either side by a total distance from one side to the other of s (for "slack") given by s^2=t^2-x^2, where t is the length of that piece and x is the distance between its end points. If you hold the end points of a two joined strings fixed and unpin the joined ends, then the total s is greater than or equal to the sum of the individual s values. The same relationships between x, t and s apply to space-time motion in special relativity.

It is meaningful within any given frame of reference; my description is in terms of what a single inertial observer sees.

True. As we were discussing the effect of a given velocity, I was assuming it was effectively constant.

12. Apr 26, 2007

### Xezlec

To me, that statement sounds like a contradiction. Are you using some odd defeinition of "accelerate"? The definition of acceleration which I am aware of is how much velocity you add per unit time. So, saying "you accelerate at X per second^2 but you don't add X per second to your velocity per second" seems like a logical contradiction, a contradiction in pure language, irrespective of any physics and mathematics involved. You're saying that P is true but the definition of P is false.

Within our frame of reference, one can certainly say, with perfect correctness, that the acceleration of the object decreases as it approaches the speed of light. It cannot possibly stay at 10 m/s^2, by the standard definition of acceleration. If we apply a constant force (using the usual definition of force) to the object, we really would find that its acceleration decreases as its velocity increases.

Now, we could choose to reconcile that with F=ma one of at least two ways: either by replacing "a" with a different definition of acceleration (in which velocity doesn't add linearly), as you might be suggesting, or by replacing "m" with "m + E/c^2" (I've proven this on paper; it's not wrong). But we must replace "m" as above because energy has inertia and must be taken into account. So, since we obviously have to do that, and since that alone makes everything work out perfectly, why replace "a"?

I'm not trying to be argumentative, I just really don't understand what's wrong with that explanation.

13. Apr 26, 2007

### Staff: Mentor

Certainly!

When we refer to "constant acceleration" in this context, we usually mean "constant proper acceleration." Although the accelerating object is not "in" a single inertial reference frame (more precisely, it does not remain at rest in any inertial reference frame), at each instant it is momentarily at rest in some inertial reference frame, corresponding to its instantaneous velocity as measured in "our" inertial reference frame. We can call this the object's "instantaneously co-moving inertial reference frame" at that instant. An object with constant proper acceleration has the same instantaneous acceleration in each of those instantaneously co-moving inertial reference frames, at the instant in each of those frames when it is momentarily at rest.

We can achieve this constant proper acceleration in a rocket ship by adjusting the thrust of the engine so as to produce a constant reading on an inertial accelerometer carried by the ship. (We have to adjust the thrust continuously to compensate for the decreasing mass of the ship as its fuel is expended.)

For a crude inertial accelerometer, the ship's pilot could use the sensation of his chair pushing against his butt.

14. Apr 26, 2007

### Xezlec

Oh, I see it now. Cool!