# Are we moving at the speed of light?

1. Mar 11, 2013

### ebodet18

Hey everyone just a quick question, I'm trying to understand E=mc2 and I keep getting conflicting information that we ARE moving at the speed of light and some saying that we are not all moving at c. Can anyone clear this up? Thanks!

2. Mar 11, 2013

### Staff: Mentor

I suspect it's because different people are using different definitions of "moving". It would be easier to tell if you gave some more details on the conflicting information you are receiving, but briefly:

(1) If by "moving" we mean "moving through space", no object with nonzero rest mass (i.e., no piece of ordinary matter like you and I are made of) can move through space at the speed of light; any such object can only move through space slower than light.

(2) If by "moving" we mean "moving through spacetime", then it can be said that an object with nonzero rest mass is moving through spacetime at the speed of light. But you have to be careful drawing deductions from that statement, which is why many people (including me) don't like that way of putting it.

Btw, how is this related to you trying to understand E = mc^2?

3. Mar 11, 2013

### ebodet18

Well for me to understand E=mc2 I first like to know everything about its part individuality (maybe some weird ocd thing) but anyway so you're saying that when moving thru spacetime 4d we all move at the speed of light, but when moving thru space 3d we don't move at the speed of light?

4. Mar 11, 2013

### Staff: Mentor

Fair enough, but I'm not sure that the whole "moving at c" thing is going to help much with this, since it doesn't have much to do with what "c" means, it's more about what "moving" means.

Kind of, but this way of putting it may lead to confusion. I didn't mean to imply that "moving through spacetime" and "moving through space" are different things. They're not; they're just different ways of looking at the same thing. More precisely, every object with nonzero rest mass is always moving through spacetime at the speed of light (with a suitable definition of "moving through spacetime"--again, many people, including me, don't like this terminology because it can invite confusion); but it may or may not be moving through space, depending on which reference frame you choose. In the object's own rest frame, it isn't moving through space; in any other frame, it is, but always at a speed slower than the speed of light.

5. Mar 11, 2013

### PAllen

Moving at the speed of light through spacetime is just a way of looking at the fact that speed and time dilation factor of a body in any frame are related such that you can write an expression of them that is always c. Then you can choose to call this expression speed through spacetime. But this quantity doesn't correspond to any measurement or observable. That is why other people don't call this speed through spacetime - they just call it the norm of 4-velocity using a common convention (this includes me; I find this whole terminology - speed through spacetime - more of a gimmick than a meaningful concept).

The expression is: √ (v^2 + c^2*d^2) = c = speed through spacetime
where d is the time dilation factor of the body in that frame, and v is the speed in that frame. Then, using this artifice, you call c*d 'speed through time', with v being speed through space. The basis for calling c*d speed through time is that it can be written c(d$\tau$/dt), and d$\tau$/dt is the rate of change of a clock moving with the body relative to a colocated clock stationary in the frame.

Last edited: Mar 11, 2013
6. Mar 11, 2013

### Pyzyq

Minute physics has a good insight of what im going to explain to you. E=mc2 describes that energy and mass are equal. This only includes and applys to objects with mass and states that those objects can only move up to afactor of the speed of light. Squared is just the unit of meassurement. The full equation is e = mc2 + pc2 now lets break it down. The e equals mc squared still means the same thing but what about the pc. Well just use the pathagry in therum, a squared plus b squared equals c squared to get the answer. E=pc means the energy (or mass) of a massless particle c has a momentum upto a factor of the speed of light and cant go no more nor no less. I suggest watchingthe vid on utube e = mc squared is not the whole story by minute physics

7. Mar 11, 2013

### Staff: Mentor

No, it isn't. It's this:

$$E^2 = m^2 c^4 + p^2 c^2$$

As you note, if m = 0 (such as for a photon), then $E = pc$. But for a particle with nonzero m, both terms can be nonzero if the particle is moving. (If the particle is not moving, then the above equation reduces to our old friend $E = m c^2$.)

8. Mar 11, 2013

### bobc2

There is an interpretation of special relativity that views the universe as 4-dimensional. Some refer to this universe model as the "block universe." You can google it for more detail. Or I can provide sketches and more discussion if you are interested. Here is a sketch from Paul Davies's book, "About Time." It identifies different ways of understanding time. It is often claimed that Einstein embraced the block universe concept. Davies and others make that claim, and Davies claims that most physicists accept the block universe concept. Certainly his close friend and colleague, Hermann Weyl did, who pictured all objects in the universe as 4-dimensional objects, including our human bodies.

Thus, all objects would be static--in effect frozen motionless in 4-dimensional space. Then how could anything be moving? Answer: No physical objects move at all. Weyl described the situation as the consciousness moving along the 4-dimensional brain of a motionless 4-dimensional body. The consciousness moves at the speed of light along the bundle of neuron fibers, strung out along the 4th dimension for typically billions of miles. It would be kind of analogous to a film strip unwould from its reel, strung out for some extended distance. You move along, your eye focused through a magnifying glass, passing each frame at a rate of around 20 frames per second, viewing the sequence of frames, conveying the impression of evolving visible activity. The whole film strip, all events in time, are all there at once, but your consciousness views the frames in rapid sequence.

Last edited: Mar 11, 2013
9. Mar 11, 2013

### 1977ub

10. Mar 12, 2013

### bobc2

Epstein has observers moving along their own worldlines at the speed of light. But, the Loedel diagrams make much more sense. At first glance at a couple of Epstein's diagrams showed one observer moving faster than the speed of light (the time axis of the second observer was rotated more than 45-degrees. And I never did see a diagram with the worldline of a photon shown to clarify that confusion. There was nothing I could see in the diagrams to indicate both observers measured the same value for the speed of light, although I may not have studied them closely enough. It was advertised as relativity made simple--I don't see that.

11. Mar 13, 2013

### Staff: Mentor

Some individuals, myself included, like to envision the fundamental geometry of 4D spacetime as including time as an entity on par with the other three spatial directions. In such a framework, a differential position vector $\vec{ds}$ in spacetime between two neighboring events at t,x,y,z and t +dt, x +dx, y +dy, z +dz can be expressed mathematically in component form by:
$$\vec{ds}=\vec{i_t}cdt+\vec{i_x}dx+\vec{i_y}dy+\vec{i_z}dz$$
where the i's are unit vectors in the spatial directions.
This is analogous to how we express a vector joining two neighboring points in 3D space. But where is this mysterious 4th time direction that is implied here? Why can't we see into that direction? In any inertial frame of reference, we have complete access to only a specific 3D cut out of 4D hyperspace (and can potentially see infinitely far into any of these three dimensions). The 4th dimension (time direction) is oriented perpendicular to our 3D cut, and, because of our inherent physical limitations as 3D beings, we have no direct access or vision into this 4th spatial dimension.

By analogy, we are like 2 dimensional beings trapped within a flat plane that is immersed in a 3D space. We have no access to the 3rd dimension, except for the 2D cross section that we currently occupy. This 2D cross section may not be stationary in 3D space; it can be moving forward (unbeknownst to us) into the 3rd dimension. If so, as time progresses, we would be sweeping out all of 3D space, and would ultimately be able to sample all of 3D space with our planar cross section. However, at any one instant of time, we would only have access to a single planar slice out of 3D space.

This is analogous to what we are experiencing in 4D hyperspace. We are 3D beings trapped within a specific 3D slice out of 4D spacetime. This 3D slice is unique to the particular inertial reference frame we currently occupy (i.e., our rest frame). We have no access or vision into our own 4th dimension, except for this 3D cut. The cut is not stationary; it is moving forward (unbeknownst to us) into our 4th spatial dimension (at the speed of light, see below). As time (measured by the synchronized clocks in our 3D reference frame) progresses, we are sweeping out all of 4D spacetime, and will ultimately be able to sample all of 4D spacetime with our moving 3D cut. However, at any one instant of time, we only have access to a single 3D cut out of 4D spacetime (a 3D panoramic snapshot). Finally, different inertial reference frames in relative motion possess different 3D cuts, and different time directions perpendicular to the 3D cuts.

If two events occur at the same spatial location within our rest frame of reference, then, according to the equation above, the differential position vector in 4D spacetime between these two events is given by:
$$\vec{ds}=\vec{i_t}cdt$$
This implies that, although, as reckoned from our inertial frame of reference, we believe we are at rest, we actually are not at rest relative to 4D spacetime. According to the present interpretation of 4D spacetime geometry, we are actually covering ground in our own time direction with a velocity of $\frac{\vec{ds}}{dt}=c\vec{i_t}$; clocks in our rest frame are not only clocks, they are also odometers for the distance we cover in spacetime.

Of course, there is a significant geometric difference between 4D spacetime and a 4D Euclidean space. In a 4D Euclidean space, if we dotted the vector $\vec{ds}$ with itself, we would obtain:
$$(ds)^2=(cdt)^2+(dx)^2+(dy)^2+(dz)^2$$
In 4D (non-Euclidean) spacetime, we have:
$$(ds)^2=-(cdt)^2+(dx)^2+(dy)^2+(dz)^2$$
This implies that in the present geometric interpretation of 4D spacetime, the dot product of the unit vector in the time direction with itself is -1, rather than + 1. This is the key geometric difference between 4D Euclidean space and 4D (non-Euclidean spacetime).

12. Mar 13, 2013

### WannabeNewton

Where have you seen such a quantity defined for arbitrary space - times i.e. in what textbook?

13. Mar 13, 2013

### chill_factor

you are currently moving at 0.99c in some reference frame.

14. Mar 13, 2013

### Prima Terra

I too have a question on the equation E=MC2. The terminology is this...energy equals mass times the speed of light. Squared. Now, it is generally accepted that there is no speed faster than light. However in this equation, M must represent a number, lets call it 2. That would mean...energy = 2x the speed of light. Now if I multiply the theory of anything by 2, then the result is greater than if I multiplied it by 1. since nothing is greater than 1x the speed of light (for the speed of light is constant) does that mean that in every instance of the equation m must represent the number 1? Or was Einstein a real evil genius...

15. Mar 13, 2013

### chill_factor

E=mc^2 is dimensionally correct. it has units of energy. mutliplying everything by 2 is just changing units. I can make the speed of light any finite number I want with the right units.

16. Mar 13, 2013

### Staff: Mentor

No, of course not. And it's not true that "nothing is greater than the speed of light". The number 2c is greater than c. So what? (What is true is that the speed of any massive object is always less than c. But that's not directly relevant to this equation.)

17. Mar 13, 2013

### Prima Terra

I do believe, that the speed of any object, small, medium or large, will always be less than c. That is ALL mass'. So nothing, so far as man has measured, is faster than the speed of light. I understand the theory of warp mechanics. However we must remain at our limits. The number 1 is e perfect representation of the speed of light. (twas Einsteins way)...as in, c cannot be 1.2, c is the speed of light. That would be1.2xc..or a little bit faster than the speed of light. We must remain within our measurable limits.

18. Mar 13, 2013

### Staff: Mentor

True, but irrelevant to understanding E = mc2.
There are systems of units that use c = 1. But standard units work just fine. If you measure mass in kg, speed in m/s, then the associated rest energy will be given in Joules.

19. Mar 13, 2013

### Prima Terra

Aye, but you have to read the original question, the young man asked if E=MC2 means we are all moving at the speed of light. The answer of course is no. The basics of the equation are representative. The 'mass' as in 'one' atom, is unmeasurable within the reaction, so it is represented by the number 1. (It is assumed to whole up to the moment of fissure) The speed of light, (186,282 miles per second) is represented by the number 1, as in 1x(186,282mph). So the base equation, in relation to Einsteins quandry is...E=M(1)xC(1)2. Now why is it squared...because of what it is trying to measure...or simply it is a measurement. Now if you open up any random internet search bar like so many non-physicists do these days and type in "what is the speed of light"...you will get this reply (or there and thereabouts)...

"One of the revelations of Albert Einstein's theory of relativity was that the speed of light in a vacuum was a constant and that nothing could travel faster."

Http://space.about.com/od/astronomybasics/a/The-Speed-Of-Light.htm....this [Broken] is the ip address as it was displayed. Its from basic quests as these that a lot of people start to formulate an interest in physics.

Last edited by a moderator: May 6, 2017
20. Mar 13, 2013

### Staff: Mentor

So far, so good. You should probably have stopped here.
No, neither the mass nor the speed of light is represented by the number 1.