Are we moving at the speed of light?

[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!

 

[PeterDonis]

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?

 

[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?

 

[PeterDonis]

Well for me to understand E=mc² I first like to know everything about its part individuality

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.

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?

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.

 

[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.

 

[Pyzyq]

Minute physics has a good insight of what I am 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 let's 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 can't go no more nor no less. I suggest watchingthe vid on utube e = mc squared is not the whole story by minute physics

 

[PeterDonis]

The full equation is e = mc2 + pc2

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 $E = m c^2$.)

 

[bobc2]

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!

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.

 

[1977ub]

This approach: http://www.relativity.li/en/epstein2/read/c0_en/c1_en/

 

[bobc2]

This approach: http://www.relativity.li/en/epstein2/read/c0_en/c1_en/

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.

 

[Chestermiller]

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 anyone 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 anyone 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 possesses 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).

 

[WannabeNewton]

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

 

[chill_factor]

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

 

[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, let's 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...

 

[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.

 

[Doc Al]

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, let's 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?

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.)

 

[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.

 

[Doc Al]

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.

True, but irrelevant to understanding E = mc².

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.

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.

 

[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 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.

 

[Doc Al]

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.

So far, so good. You should probably have stopped here.

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).

No, neither the mass nor the speed of light is represented by the number 1.

 

[Chestermiller]

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

Wills, A. P., Vector Analysis With and Introduction to Tensor Analysis, Dover, 1958, Chapter IX, Non-Euclidean Manifolds

"For the purposes of what may be called pre-relativity physics a vector analysis dealing with vectors (and tensors) of our 3-dimensional Euclidean space was fairly adequate. But modern advances in physics, exemplified particularly by relativity theory, demanded an extension in non-Euclidean space of four dimensions ..."

 

[PAllen]

Wills, A. P., Vector Analysis With and Introduction to Tensor Analysis, Dover, 1958, Chapter IX, Non-Euclidean Manifolds

"For the purposes of what may be called pre-relativity physics a vector analysis dealing with vectors (and tensors) of our 3-dimensional Euclidean space was fairly adequate. But modern advances in physics, exemplified particularly by relativity theory, demanded an extension in non-Euclidean space of four dimensions ..."

I think what was being asked about was the ict construction. That really only works for flat spacetime, with standard coordinates.

 

[WannabeNewton]

I think what was being asked about was the ict construction. That really only works for flat spacetime, with standard coordinates.

Yes that's what I was talking about. That quantity doesn't make any sense for arbitrary space times.

 

[bobc2]

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 anyone 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 anyone 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 possesses 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).

Very nice summary, Chestermiller. Thanks for the details. I think you have shown us a very instructive way of understanding it.

 

[jtbell]

Wills, A. P., Vector Analysis With and Introduction to Tensor Analysis, Dover, 1958, Chapter IX, Non-Euclidean Manifolds

Wills died in 1937, according to Wikipedia. That book was actually first published in 1931 (apparently):

http://books.google.com/books/about/Vector_analysis.html?id=5Nk-AAAAIAAJ

(scroll down to "Other Editions")

 

[Chestermiller]

Yes that's what I was talking about. That quantity doesn't make any sense for arbitrary space times.

No. The development in Wills did not ever invoke the ict construction or even mention it. Why don't you guys read it before you go criticizing it. The Dover publications have always been written by specially selected and very highly regarded authors. Also, jtbell, just because Wills died in the 1930's does not mean that his book has not stood the test of time. Don't criticize him by innuendo. If you have any specific criticisms about the correctness of his mathematics or its applicability to general relativity, let's hear them.

Chet

 

[bobc2]

No. The development in Wills did not ever invoke the ict construction or even mention it. Why don't you guys read it before you go criticizing it. The Dover publications have always been written by specially selected and very highly regarded authors. Also, jtbell, just because Wills died in the 1930's does not mean that his book has not stood the test of time. Don't criticize him by innuendo. If you have any specific criticisms about the correctness of his mathematics or its applicability to general relativity, let's hear them.

Chet

There is certainly nothing wrong with Wills's math, nor yours, Chestermiller. It's not as though either of you invented mathematics. Do you suppose we should reconsider Newton's calculus, since it is so old, you know.

 

[WannabeNewton]

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:
$$\vec{ds}=\vec{i_t}cdt+\vec{i_x}dx+\vec{i_y}dy+\vec{i_z}dz$$

Ok then. Would you like to show how this object makes sense for ANY arbitrary space - times as you have claimed? Vectors at a point lie in a single tangent space and you must parallel transport them, via the affine connection (in GR the levi - civita connection), to other tangent spaces to compare them. What you did only works in flat space - time so already this quantity you have written down makes no sense in full generality (a vector "between" two events). Secondly, $ds^2$ is NOT the result of a dot product. If $(M,g_{ab})$ is a space - time and $p\in M$ then we have a chart $(U,\varphi )$ such that the coordinate domain contains $p$. Using the coordinate basis vector fields, $\left \{ \partial _{1},...,\partial _{4} \right \}$ defined on this coordinate domain, we write $g = g(\partial _{i},\partial _{j})dx^{i}\otimes dx^{j}$ where $dx^{a}$ are the dual one - form fields and it is really this quantity you are talking about when writing down $ds^2$. You can't just go whilly nilly and do things the way they work in flat space - time because in flat space - time you can get away with many hand wavy arguments but not so in curved space -time.

 

[AnTiFreeze3]

... Also, jtbell, just because Wills died in the 1930's does not mean that his book has not stood the test of time. Don't criticize him by innuendo ...

Really? No one criticized the author. Jtbell simply found a discrepancy between your supplied publication date and the year in which Wills died. It's entirely plausible that Dover reissued another edition after the death of Wills, but it makes absolutely no sense to think that anyone was criticizing the book you use, or its author.

 

[Chestermiller]

Ok then. Would you like to show how this object makes sense for ANY arbitrary space - times as you have claimed? Vectors at a point lie in a single tangent space and you must parallel transport them, via the affine connection (in GR the levi - civita connection), to other tangent spaces to compare them. What you did only works in flat space - time so already this quantity you have written down makes no sense in full generality (a vector "between" two events). Secondly, $ds^2$ is NOT the result of a dot product. If $(M,g_{ab})$ is a space - time and $p\in M$ then we have a chart $(U,\varphi )$ such that the coordinate domain contains $p$. Using the coordinate basis vector fields, $\left \{ \partial _{1},...,\partial _{4} \right \}$ defined on this coordinate domain, we write $g = g(\partial _{i},\partial _{j})dx^{i}\otimes dx^{j}$ where $dx^{a}$ are the dual one - form fields and it is really this quantity you are talking about when writing down $ds^2$. You can't just go whilly nilly and do things the way they work in flat space - time because in flat space - time you can get away with many hand wavy arguments but not so in curved space -time.

Dear WannabeNewton,

When I wrote down that equation, I was trying to keep things simple for the OP. Yes, of course the equation as I wrote it does not apply in curved space or even when using curvilinear coordinates in flat space. The extension of this starting point to curved tangent spaces in a way that satisfies all the requirements you cited above are laid out in full detail in Wills' book. I suggest you get a copy of the book, and see what he actually wrote. I did not intend to give the impression that the equation that I wrote was applicable in general relativity.

I'm not going to go through the details of Wills' development here, because it is too lengthy to be appropriate for this venue. But, to stimulate your interest, I will give the starting point in the development which is the expression for a differential position vector:

$$\vec{ds}=\vec{a_j}dx^j$$

where the $\vec{a_j}$'s are the coordinate basis vectors, and the dx^j's are the differentials in the coordinates. If we take the dot product of the vector $\vec{ds}$ with itself, we get:
$$(ds)^2=g_{ij}dx^idx^j$$
where the g_ij are the components of the metric tensor:

$$g_{ij}=\vec{a_i}\cdot \vec{a_j}$$

Here is a specific example of how this applies to the Schwartzschild metric:
$$\vec{ds}=\vec{a_t}cdt+\vec{a_r}dr+\vec{a_\theta}d \theta+\vec{a_\phi}d\phi$$
with $$\vec{a_t}=\sqrt{1-2 \frac{m}{r}}\vec{i_t}$$
$$\vec{a_r}=\frac{1}{\sqrt{1-2 \frac{m}{r}}}\vec{i_r}$$
$$\vec{a_\theta}=r\vec{i_\theta}$$
$$\vec{a_\phi}=rsin\theta \vec{i_\phi}$$

I hope this stimulates your interest. It is not exactly the way you learned it, but that doesn't mean that it is incorrect. Please try to keep an open mind. For some people, the development expressed in this physically motivated way simplifies things considerably.

Chet