Discover the Relationship Between E=mc2 and Time with m=1kg and d=1m

[Deepak K Kapur]

If i put c=d/t in
E=mc2, then E=m×d2/t2

Now take m=1kg and d=1m

Does this mean that E is inversely proportional to time?

 

[Fredrik]

If i put c=d/t in
E=mc2, then E=m×d2/t2

Now take m=1kg and d=1m

Does this mean that E is inversely proportional to time?

No, that would be the case if there's a constant A such that E=A/t. If there's a constant A such that E=A/t², that we could say that E is inversely proportional to time squared. But what you found is E= md²/t², where m is a constant and d depends on t. So md² isn't a constant.

 

[PeroK]

If i put c=d/t in
E=mc2, then E=m×d2/t2

Now take m=1kg and d=1m

Does this mean that E is inversely proportional to time?

No, but it means that Energy has a "dimension" of $ML^2T^{-2}$.

Compare this with the classical equation $KE = \frac12 mv^2$, where kinetic energy has the same dimension as above.

Look up "dimensional analysis".

 

[Deepak K Kapur]

No, that would be the case if there's a constant A such that E=A/t. If there's a constant A such that E=A/t², that we could say that E is inversely proportional to time squared. But what you found is E= md²/t², where m is a constant and d depends on t. So md² isn't a constant.

Okay, i get your point.

But,
1. d=1m and one meter is always one meter. How does 1m depend on time?

2. When d=1m and m=1kg, we get E=1/t2. What is the physical significance of this reduced equation?

Thanks.

 

[Deepak K Kapur]

No, but it means that Energy has a "dimension" of $ML^2T^{-2}$.

Compare this with the classical equation $KE = \frac12 mv^2$, where kinetic energy has the same dimension as above.

Look up "dimensional analysis".

Thanks for your answer.

One more question (silly one).

Mass is something concrete, we can see it, touch it, feel it etc.
Speed is something abstract, we can't see it, touch it etc. apart from mass. It doesn't seem to have independent existence.

Then, how on Earth can we multiply a concrete entity with an abstract one??

 

[Fredrik]

1. d=1m and one meter is always one meter. How does 1m depend on time?

1 m doesn't depend on time, but this d is defined by d=ct. If it's not, then we're no longer talking about velocity c. So d "depends on" t in the sense that the number represented by the variable d can be calculated if you know the number represented by the variable t.

2. When d=1m and m=1kg, we get E=1/t2.

When d=1 m, then t is (1 m)/(299792458 m/s), so what we get is
$$E=(1~\mathrm{kg}) \frac{(1~\mathrm{m})^2}{\left(\frac{1~\mathrm{m}}{(299792458~\mathrm{m/s}}\right)^2} = (1~\mathrm{kg}) (299792458~\mathrm{m/s})^2 =(299792458)^2~\mathrm{kgm^2/s^2} =(299792458)^2~\mathrm{J}.$$

 

[Dale]

Mass is something concrete, we can see it, touch it, feel it etc.
Speed is something abstract, we can't see it, touch it etc. apart from mass. It doesn't seem to have independent existence.

You are confusing mass with matter. You can touch and feel matter. Mass is a property that systems of matter have. It also does not have an independent existence.

Besides, there is no reason that the operation of multiplication requires "abstract" or "concrete" entities to be multiplied by themselves. The concepts of "abstract" and "concrete" don't even enter into the operation.

 

[Deepak K Kapur]

1 m doesn't depend on time, but this d is defined by d=ct. If it's not, then we're no longer talking about velocity c. So d "depends on" t in the sense that the number represented by the variable d can be calculated if you know the number represented by the variable t.When d=1 m, then t is (1 m)/(299792458 m/s), so what we get is
$$E=(1~\mathrm{kg}) \frac{(1~\mathrm{m})^2}{\left(\frac{1~\mathrm{m}}{(299792458~\mathrm{m/s}}\right)^2} = (1~\mathrm{kg}) (299792458~\mathrm{m/s})^2 =(299792458)^2~\mathrm{kgm^2/s^2} =(299792458)^2~\mathrm{J}.$$

So, it means that in this equation no kind of relation between energy and time can be ever found out...is it so??

 

[Deepak K Kapur]

You are confusing mass with matter. You can touch and feel matter. Mass is a property that systems of matter have. It also does not have an independent existence.

Besides, there is no reason that the operation of multiplication requires "abstract" or "concrete" entities to be multiplied by themselves. The concepts of "abstract" and "concrete" don't even enter into the operation.

Does it then mean that when we do mathematics ( which is something abstract), then every kind of entity, whether concrete or abstract enters the 'abstract' realm?

 

[Dale]

Mathematical operations are certainly abstract, whether the quantities represented by the math are abstract or not. However, "enters the 'abstract' realm" sounds more like something that you will find in a fantasy novel than in a scientific textbook, so I don't have a direct answer to your question.

 

[Fredrik]

So, it means that in this equation no kind of relation between energy and time can be ever found out...is it so??

Not necessarily. But it means that most candidates for such relations can be immediately discarded because they contradict experiments that don't contradict relativity.

Does it then mean that when we do mathematics ( which is something abstract), then every kind of entity, whether concrete or abstract enters the 'abstract' realm?

I would say that aspects of the real world are represented by abstract mathematical things in the theory.

 

[russ_watters]

2. When d=1m and m=1kg, we get E=1/t2.

No we don't. The units do not disappear.

 

[Deepak K Kapur]

You are confusing mass with matter. You can touch and feel matter. Mass is a property that systems of matter have. It also does not have an independent existence.

So, can it be said that when we do science or math, we are not concerned with the 'real' things but are only bothered about their properties??

 

[Deepak K Kapur]

No we don't. The units do not disappear.

Cant we say in simple language that when a mass of 1kg moves a distance of 1m then as per E=mc2, we have E=1/t2.

Why not, this seems to be a sensible interpretation...

 

[ShayanJ]

Cant we say in simple language that when a mass of 1kg moves a distance of 1m then as per E=mc2, we have E=1/t2.

Why not, this seems to be a sensible interpretation...

t can not be a variable. The point is, $c=299792458 \ \frac m s=\frac{299792458 \ m}{1 \ s}=\frac{149896229 \ m}{.5 \ s}=\frac{59958491.6 \ m}{.2 \ s}=\dots$

So you can only have $E=m(\frac{299792458}{1})^2=m(\frac{149896229}{.5})^2=m(\frac{59958491.6}{.2})^2=\dots$.

You're just substituting different representations of the same number c in the formula $E=mc^2$! Its like getting the formula $E_k=\frac 1 2 mv^2$ and writing it as $E_k=\frac{16}{32} mv^{\sqrt{4}}$!

EDIT:
More clearly, if you take d=1 m, then your t is 1/c. So the representation you're using is $E=m(\frac{1}{\frac{1}{c}})^2=m(\frac{1}{\frac{1}{299792458}})^2 \approx m(\frac{1}{3.33 \times 10^{-9}})^2$!

 

[russ_watters]

Cant we say in simple language that when a mass of 1kg moves a distance of 1m then as per E=mc2, we have E=1/t2.

Why not, this seems to be a sensible interpretation...

Well, two reasons:
1. Because the units are still there and you are pretending they aren't. The "1" has units kg-m2. The equation should read E=1 kg-m2/t2.
2. The equation isn't describing the motion of a kilogram of mass, the piece you broke apart is describing the speed of light ("c" is the speed of light). So the 1m is the distance traveled by light in "t" time: time is a constant here, not a variable and you can't change that. So when you plug in m=1 kg, and d= 1m, t=1/300,000,000s and it all simplifies to E=9x10^16 Joules.

So, can it be said that when we do science or math, we are not concerned with the 'real' things but are only bothered about their properties??

Science includes the implied assumption that the observed properties are real and there is nothing else to "concern with" or not "concern with". Beyond that, you're getting into philosophy, not science.

 

[Deepak K Kapur]

Well, two reasons:
1. Because the units are still there and you are pretending they aren't. The "1" has units kg-m2. The equation should read E=1 kg-m2/t2.
2. The equation isn't describing the motion of a kilogram of mass, the piece you broke apart is describing the speed of light ("c" is the speed of light). So the 1m is the distance traveled by light in "t" time: time is a constant here, not a variable and you can't change that. So when you plug in m=1 kg, and d= 1m, t=1/300,000,000s and it all simplifies to E=9x10^16 Joules.

Science includes the implied assumption that the observed properties are real and there is nothing else to "concern with" or not "concern with". Beyond that, you're getting into philosophy, not science.

I don't get your point fully...

1. 'c' in this equation implies that motion of mass is involved. So, why not describe the motion of a kg of mass?

2. How is time constant here? If i put t= 2 sec, the distance gets doubled to keep the speed of light constant. So, how is time constant?

 

[ArmanCham]

Time constant means this ;
c=x/t If we assume x is one then you cannot change t cause c is constant.So you are telling us E=1/t² here you assumed x equal one.Then you assumed t =2 We cannot change time If you choose x =1.If you change time you can see easily that you are breaking fundamental physics law "Speed of light is constant"
Here the math c=x/t you said x =1 then t=1/c.Now we found t.Its a number but you are telling it can be three or four.Thats nonsense.

 

[russ_watters]

1. 'c' in this equation implies that motion of mass is involved.

You are mistaken. C is just a universal constant, that happens to be the speed of light.

2. How is time constant here? If i put t= 2 sec, the distance gets doubled to keep the speed of light constant. So, how is time constant?

You stated the distance was 1m, which just makes time the new constant in the equation. No matter how you change the distance, the time must change in the opposite way in order that d/t always equals C. That's why plugging-in C=d/t is just an unnecessary complication: C is always going to be the same, so there is no point in calculating it every time you do the problem.

You really need to stop trying to break the equation. If you get to a point where you think you have, all it means is that you've confused yourself.

 

[Deepak K Kapur]

You are mistaken. C is just a universal constant, that happens to be the speed of light.

You stated the distance was 1m, which just makes time the new constant in the equation. No matter how you change the distance, the time must change in the opposite way in order that d/t always equals C. That's why plugging-in C=d/t is just an unnecessary complication: C is always going to be the same, so there is no point in calculating it every time you do the problem.

You really need to stop trying to break the equation. If you get to a point where you think you have, all it means is that you've confused yourself.

I will not argue further on the time issue...
But,
1. if v in KE=1/2mv2 implies the motion of mass, why doesn't c in E=mc2 imply the motion of mass.

2. What does c in this equation mean?

Thanks, u hav been very helpful...

 

[russ_watters]

1. if v in KE=1/2mv2 implies the motion of mass, why doesn't c in E=mc2 imply the motion of mass.

There is no "implies" here: V is the speed of the object, C is the speed of light. Though the two equations bear some similarities, they are not the same equation.

2. What does c in this equation mean?

Again, c is the speed of light. Or do you really mean what does the equation mean? The equation provides a conversion that demonstrates the equivalence of matter and energy. So, for example, if you weigh the fuel of a nuclear reactor after it is spent and find it to be less than its starting weight, this equation tells you how much energy was released by converting the lost matter into energy.

http://en.m.wikipedia.org/wiki/Mass–energy_equivalence

 

[Fredrik]

1. if KE=1/2mv2 implies the motion of mass, why doesn't c in E=mc2 imply the motion of mass.

In non-relativistic classical mechanics, the total energy of a free particle of mass m moving at speed v is $\frac 1 2 mv^2$. In special relativity, it's $\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$. Note that this is equal to $mc^2$ if and only if $v=0$.

 

[Deepak K Kapur]

Again, c is the speed of light. Or do you really mean what does the equation mean? The equation provides a conversion that demonstrates the equivalence of matter and energy. So, for example, if you weigh the fuel of a nuclear reactor after it is spent and find it to be less than its starting weight, this equation tells you how much energy was released by converting the lost matter into energy.

http://en.m.wikipedia.org/wiki/Mass–energy_equivalence

Actually i was expecting the meaning of this equation as follows..

Matter will change into energy when......(some relation to the speed of light)
Cant you elaborate this equation in this way?

Also please explain the meaning of 'square' of c?

 

[Deepak K Kapur]

In non-relativistic classical mechanics, the total energy of a free particle of mass m moving at speed v is $\frac 1 2 mv^2$. In special relativity, it's $\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$. Note that this is equal to $mc^2$ if and only if $v=0$.

Also, i read on the net that one can have c=1 also. If this is the case, what meaning is left in 'conversion' then. If c=1, does it mean that energy and mass are same?

If they are same, then do they look different to the 'observer' only or are they 'really' different?

 

[russ_watters]

Actually i was expecting the meaning of this equation as follows..

Matter will change into energy when......(some relation to the speed of light)
Cant you elaborate this equation in this way?

No, that isn't what it means. I can't really elaborate on something that isn't true.

Also please explain the meaning of 'square' of c?

C^2 is the conversion factor to equate matter and energy.

 

[russ_watters]

Also, i read on the net that one can have c=1 also. If this is the case, what meaning is left in 'conversion' then. If c=1, does it mean that energy and mass are same?

If they are same, then do they look different to the 'observer' only or are they 'really' different?

Not the same, equivalent.

Note, the actual value of C depends on what units you express it in. It can be anything as long as those units are consistent throughout the equation.

 

[Deepak K Kapur]

Not the same, equivalent.

Would you like to elaborate?

 

[russ_watters]

Would you like to elaborate?

Equivalent in value, but not the same in subtance (if that word even applies). IE, two apples vs two oranges.

Please read the wiki link I provided.

 

[Deepak K Kapur]

I will not argue further on the time issue...
But,
1. if v in KE=1/2mv2 implies the motion of mass, why doesn't c in E=mc2 imply the motion of mass.

2. What does c in this equation mean?

Thanks, u hav been very helpful...

Somehow, i can't let it go...please bear with me

1. In KE=1/2m × v2, if we replace the multiplication sign '×' by the addition sign '+' (just suppose), we get, KE=1/2m +v2.

In other words '×' means that the object'moves' with velocity 'v'
whereas
'+' does not mean so.

2. So, why does '×' in E=m x c2 does not mean that the object moves with c (or near c), even if 'c' is the speed of light. It's speed afterall...

 

[jbriggs444]

1. In KE=1/2m × v2, if we replace the multiplication sign '×' by the addition sign '+' (just suppose), we get, KE=1/2m +v2.

This replaces an equation that is motivated by centuries of experiment and replaces it with something that was conjured up with the stroke of a pen and which is not even dimensionally consistent.

In other words '×' means that the object'moves' with velocity 'v'
whereas
'+' does not mean so.

That is not what times and plus mean.

 

[Vanadium 50]

Somehow, i can't let it go...please bear with me

First, enough with the text speak. It's against PF Rulez...er...Rules. You've been warned before, and the fact that you continue to do this shows a disrepect for the forum membership - the same people you are asking for help. You can do better than this.

Second, you can't just start with an advanced topic like relativity until you understand the foundations it is built on. You clearly have gaps in your understanding going back to at least Newtonian mechanics and possibly arithmetic. You'll need to fill those gaps before you can understand the ideas that build upon them.

 

[russ_watters]

Somehow, i can't let it go...please bear with me

1. In KE=1/2m × v2, if we replace the multiplication sign '×' by the addition sign '+' (just suppose), we get, KE=1/2m +v2.

In other words '×' means that the object'moves' with velocity 'v'
whereas
'+' does not mean so.

I'm sorry, but since the units don't match, that's just mathematical gibberish. I don't mean to sound insulting here, but how much schooling have you had? What is the highest level of math and science you've completed? In the US, we typically start with a course at age 14 that teaches the basic tools of science, including dimensional analysis. I've you've passed that level, you'll need to go back and brush-up on it. Here's a link discussing it:
http://www.chem.tamu.edu/class/fyp/mathrev/mr-da.html

2. So, why does '×' in E=m x c2 does not mean that the object moves with c (or near c), even if 'c' is the speed of light. It's speed afterall...

Because it was derived to mean something else. Did you read the wiki link I posted? I guess it's unfortunate that the proportionality constant happens to be C (squared), but that's all it is in this equation: a proportinality constant. C happens to be the speed of light, the maximum speed of objects (asymptotic) and a proportionality constant relating mass and energy. It can and does served different purposes in different equations. You're not alone here, it does cause a lot of confusion. IE:

Einstein's famous equation tells us that mass moving at near light speed is essentially energy.

I'm sorry, but no, it really doesn't. As already said by someone else, e=mc² is valid only for stationary objects.

It's not exactly the same, but you might think of it as being similar to how both torque and work have the same units but mean two very different things.

 

[russ_watters]

So I googled the question for other ways to say the same thing and found this relevant quote from a 13 year old PF thread:

...we know, from relativity that energy and mass are proportional. That is, Energy is some constant times mass. We know, by comparing units (dimensional analysis) that the constant must have units of "speed". And the only fixed speed in the universe is the speed of light. That's very much a "hand waving" explanation but a more accurate explanation would have to be how the equation was derived in the first place and a link to that has already been given.

And:

If you have some proportionality between energy and mass, E=bm, then just by looking at the units you know that b needs to have units of J/kg=m²/s². The only combination of fundamental constants that has those units is c², so it has to be c².

Of course, that begs the question, why is it 1 c² instead of 5 c²? And why is it E=bm instead of E=bm²? To answer either of those you really need a full derivation.

 

[DarioC]

Hope I don't regret this: Lot of complex answers here so far.

C= does not equal distance over time except for the values that are measured when determining the speed of light. C is a specific constant in this formula and cannot be used as a variable as you are trying to do because it is not true for any other values.

E=MC squared says that if you convert a certain mass to energy the amount of energy will be the mass times the speed of light squared, not d over t squared, except for the one ratio that gives the speed of light. A little bit of mass, a lot of energy.

The formula C=d/t is a specific version of V=d/t and only has one value for C, even though it can be expressed in different units that will give different units of E. (as has been said already).

You, or I, can do all kinds of things with mathematics. Sometimes the results are true and sometimes they are way wrong.

DC

 

[Deepak K Kapur]

First, enough with the text speak. It's against PF Rulez...er...Rules. You've been warned before, and the fact that you continue to do this shows a disrepect for the forum membership - the same people you are asking for help. You can do better than this.

Second, you can't just start with an advanced topic like relativity until you understand the foundations it is built on. You clearly have gaps in your understanding going back to at least Newtonian mechanics and possibly arithmetic. You'll need to fill those gaps before you can understand the ideas that build upon them.

Well...i was expecting such a response...

I am a teacher myself but never snub anyone who tries to understand something honestly, even if the quality of questions posed is dismal...

I also don't threat such a person...

Well, some powerful people tend to do such a thing...