c^2 in Einstein‘s theory of relativity

[Lightisslow]

C2 means the speed of light squared so Mass traveling at the speed of light times itself becomes energy. That would mean we could travel up to.34,500,000,000 miles per second before breaking down into pure energy. So we should be able to travel faster than the speed of life. Did I miss something?

 

[Ibix]

That makes absolutely no sense, I'm afraid. In $E=mc^2$ the $c^2$ is just a unit conversion factor between our units of mass and units of energy. You can use so-called geometric units in which $c=1$, and then the equation is just $E=m$. It tells you that mass and energy are related, but says nothing about speed limits or turning into energy.

The equation is also a special case of the more general $E^2=m^2c^4+p^2c^2$ when the momentum $p=0$. For objects with non-zero mass that more general equation simplifies to $E=\gamma mc^2$, where $\gamma=\frac 1{\sqrt{1-v^2/c^2}}$ and $v$ is the speed of the mass. If you rearrange that and solve for $v$ then this will show you that no matter how much energy you supply, $v$ will always be less than $c$.

 

[Orodruin]

C2 means the speed of light squared so Mass traveling at the speed of light times itself becomes energy.

That’s not at all what is stated. It is stated that an object has a rest energy that is its inertial mass at rest multiplied by the speed of light squared. That refers to a mass at rest.

That would mean we could travel up to.34,500,000,000 miles per second before breaking down into pure energy.

There is no such thing as ”pure energy”. There are many different forms of energy and energy conservation only states that the total remains fixed.

So we should be able to travel faster than the speed of life. Did I miss something?

You have completely made up your own theory. It has nothing to do with relativity and is also not describing reality.

 

[Dale]

That would mean we could travel up to.34,500,000,000 miles per second before breaking down into pure energy.

In addition to the other corrections that have already been posted, there is a problem with your math. Specifically the units.

$$c^2\ne 34,500,000,000 \mathrm{\ mile/second}$$$$c^2= 34,500,000,000 \mathrm{\ mile^2/second^2}$$Those are very different quantities. It is like comparing miles to acres, or apples to oranges as the saying goes. There is simply no valid way to compare two quantities with different units and say which is larger or smaller.

That has nothing to do with relativity, it is simply how units work.

 

[jtbell]

There is no such thing as ”pure energy”. There are many different forms of energy and energy conservation only states that the total remains fixed.

Another way to say this is that energy is always a property of some physical object or entity. If one object gains energy, another object must lose energy.

 

[jbriggs444]

Another way to say this is that energy is always a property of some physical object or entity. If one object gains energy, another object must lose energy.

Feynman put a nice spin on what energy is in this lecture.

 

[thomsj4]

Consider a four-dimensional spacetime with coordinates $t x y z$. For any massive particle, the interval is given by

$$ ds^2 = -c^2\,dt^2 + dx^2 + dy^2 + dz^2. $$

To move faster than light in ordinary special relativity, we would require a timelike worldline to become spacelike, meaning the interval would switch sign from negative to positive for the same particle. A hypothetical way to achieve this is to propose a continuous deformation of the metric itself: define a smooth function

$$ \Lambda(\tau) = 1 + \alpha(\tau), $$

where $\tau$ is the proper time and $\alpha(\tau)$ is a carefully constructed function such that it remains infinitesimal for almost all $\tau$ but integrates to a nontrivial topological shift over a closed loop in spacetime. Insert this $\Lambda(\tau)$ into the metric components so that each time the particle completes a topological cycle, its effective velocity limit locally looks like

$$ c_{\text{local}}(\tau) = c\sqrt{\Lambda(\tau)}. $$

If we engineer $\alpha(\tau)$ to return to zero after a finite interval of $\tau$, the geometry reverts to the original Minkowski form yet encodes a net displacement in the spacelike direction that exceeds $c\Delta t$. In other words, upon completing this “loop,” the particle has advanced farther in space than light could in the same coordinate time.

To see why this might fool us into thinking we can exceed the speed of light, notice that for each small segment of proper time $\Delta \tau$:

$$ ds^2 \approx -c^2 \bigl(1+\alpha(\tau)\bigr)\,d\tau^2 + \dots $$

so that the local speed limit depends on $\alpha(\tau)$. Summed over a whole cycle, the path length in space can accumulate faster than $c\Delta t$. In particular, the four-velocity still never violates

$$ \frac{dx^\mu}{d\tau}\,\frac{dx_\mu}{d\tau} = -c^2 $$

locally, but the global topology of the altered metric allows us to come back to the same $\tau$ while having covered more spatial distance than a simple Minkowski observer would measure as permitted in a normal inertial frame.

Mathematically, because

$$ \int_{\text{loop}} \nabla_\mu \alpha(\tau)\,dx^\mu \neq 0, $$

there is a net change in the local light cone tilt for the traveling particle, which effectively accumulates a superluminal displacement. However, this rests on the fiction of a nontrivial spacetime topology or metric deformation that we do not observe in flat Minkowski spacetime. The moment we impose standard special relativity’s geometry again (with $\alpha(\tau) = 0$ everywhere), the argument collapses, and we see there is no global inertial frame in which the particle’s velocity has exceeded $c$.

So the original assumption of interpreting $c^2$ as a speed is flawed: $c^2$ in $E = mc^2$ is a conversion factor from mass units to energy units, not a physical velocity limit. The actual limitation is due to the invariant interval $ds^2$ and the requirement that massive particles follow timelike trajectories with negative intervals:

$$ ds^2 = -c^2\,d\tau^2 < 0. $$

Thus the “speed” $c^2$ being about $34{,}500{,}000{,}000$ miles per second is never a meaningful velocity for travel, but rather an artifact of squaring $c$. This is why physically traveling faster than $c$ is still prohibited once the proper relativistic geometry is restored.

 

[Orodruin]

Consider a four-dimensional spacetime with coordinates $t x y z$. For any massive particle, the interval is given by

$$ ds^2 = -c^2\,dt^2 + dx^2 + dy^2 + dz^2. $$

To move faster than light in ordinary special relativity, we would require a timelike worldline to become spacelike, meaning the interval would switch sign from negative to positive for the same particle. A hypothetical way to achieve this is to propose a continuous deformation of the metric itself: define a smooth function

$$ \Lambda(\tau) = 1 + \alpha(\tau), $$

where $\tau$ is the proper time and $\alpha(\tau)$ is a carefully constructed function such that it remains infinitesimal for almost all $\tau$ but integrates to a nontrivial topological shift over a closed loop in spacetime. Insert this $\Lambda(\tau)$ into the metric components so that each time the particle completes a topological cycle, its effective velocity limit locally looks like

$$ c_{\text{local}}(\tau) = c\sqrt{\Lambda(\tau)}. $$

If we engineer $\alpha(\tau)$ to return to zero after a finite interval of $\tau$, the geometry reverts to the original Minkowski form yet encodes a net displacement in the spacelike direction that exceeds $c\Delta t$. In other words, upon completing this “loop,” the particle has advanced farther in space than light could in the same coordinate time.

To see why this might fool us into thinking we can exceed the speed of light, notice that for each small segment of proper time $\Delta \tau$:

$$ ds^2 \approx -c^2 \bigl(1+\alpha(\tau)\bigr)\,d\tau^2 + \dots $$

so that the local speed limit depends on $\alpha(\tau)$. Summed over a whole cycle, the path length in space can accumulate faster than $c\Delta t$. In particular, the four-velocity still never violates

$$ \frac{dx^\mu}{d\tau}\,\frac{dx_\mu}{d\tau} = -c^2 $$

locally, but the global topology of the altered metric allows us to come back to the same $\tau$ while having covered more spatial distance than a simple Minkowski observer would measure as permitted in a normal inertial frame.

Mathematically, because

$$ \int_{\text{loop}} \nabla_\mu \alpha(\tau)\,dx^\mu \neq 0, $$

there is a net change in the local light cone tilt for the traveling particle, which effectively accumulates a superluminal displacement. However, this rests on the fiction of a nontrivial spacetime topology or metric deformation that we do not observe in flat Minkowski spacetime. The moment we impose standard special relativity’s geometry again (with $\alpha(\tau) = 0$ everywhere), the argument collapses, and we see there is no global inertial frame in which the particle’s velocity has exceeded $c$.

So the original assumption of interpreting $c^2$ as a speed is flawed: $c^2$ in $E = mc^2$ is a conversion factor from mass units to energy units, not a physical velocity limit. The actual limitation is due to the invariant interval $ds^2$ and the requirement that massive particles follow timelike trajectories with negative intervals:

$$ ds^2 = -c^2\,d\tau^2 < 0. $$

Thus the “speed” $c^2$ being about $34{,}500{,}000{,}000$ miles per second is never a meaningful velocity for travel, but rather an artifact of squaring $c$. This is why physically traveling faster than $c$ is still prohibited once the proper relativistic geometry is restored.

This is a B level thread and based on the OP’s opening post I would say they have no idea what a metric is …

 

[L Drago]

That’s not at all what is stated. It is stated that an object has a rest energy that is its inertial mass at rest multiplied by the speed of light squared. That refers to a mass at rest.


There is no such thing as ”pure energy”. There are many different forms of energy and energy conservation only states that the total remains fixed.


You have completely made up your own theory. It has nothing to do with relativity and is also not describing reality.

Special relativity describes that according to equation e=mc², we cannot travel in the speed of light or faster than that because we would need infinite energy and mass to travel in the speeds of light. In the Tl;dr summary of the thread, in the last part this is mistaken and said that it is possible to travel more than speed of life ( I think it wanted to state light).

In the Tl;dr , I suppose another statement is also wrongly stated as c² is different from c and there is some problem in unit conversion. E = mc² is the energy mass equivalence and c is squared.

 

[Ibix]

I suppose another statement is also wrongly stated

It would be shorter to say what is correctly stated. I don't think he even managed to square $c$ in miles per second correctly (the third significant figure should be a seven, never mind the units).

 

[Dale]

Special relativity describes that according to equation e=mc², we cannot travel in the speed of light or faster than that because we would need infinite energy and mass to travel in the speeds of light.

I wouldn't say that. The equation $E=mc^2$ only applies when $m$ is at rest. So it doesn't say anything about travelling at any speed.

The equation that says that a massive object cannot travel at $c$ is the more general one: $$m^2 c^2 = E^2/c^2 - p^2$$ where $\vec p$ is the momentum. That combined with $\vec v=c^2 \vec p/E$ ensure that $v<c$ if $0<m$.