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Question about Photons…

  1. Feb 4, 2009 #1
    I read that "In red LEDs the electrons moving down their gradient release 1.9 electron volts in the form of photons with the same amount of energy, 1.9 electron volts."

    Is that the only way photons are made, by electrons moving down gradients and releasing them? Are they being made or are they just already part of the electron and the electron is releasing them? Or is it something else???
  2. jcsd
  3. Feb 5, 2009 #2
    There seems to be a lot of questions about photons on here!

    I think of a photons as a quantised wave. An analogy is this:

    You have a pond and a set of stones. The stones are all the same size and you can drop them from certain heights (1 cm, 3 cm, etc.). These are like electrons. The pond is like the electromagnetic field.

    When you drop the stones you will get a wave of a certain size. Whether the waves are 'part of the stone' or 'released' by the stone isn't really important.

    I hope that helps.
  4. Feb 5, 2009 #3
    I also herd that E=MC^2 can be applied to say that photons also create matter. Id like to hear a simple explanation of how they can create matter and also how the electrons make the photons...

    What exactly dose "energy equals mass times the speed of light squared" mean?

    1 pound times 700 million miles an hour squared = one coulomb of electrons??????
  5. Feb 5, 2009 #4
    Hey Cyrus, E=mc2 is probably one of the most interesting things you will come across when learning physics. The equation says that mass and energy are the same thing, two sides of the same coin. Matter can become energy and energy can become matter.

    In the situation where the electron releases the photon, there really is no conversion of matter and energy, only energy is converted into a different kind of energy. The electron gains energy from some outside source enabling it to jump to a higher energy level. The electron can only maintain this state for a short period of time before it falls back to the previous energy level, releasing the energy it had gained in the form of a photon.

    Now there are some situations where E=mc2 become easily noticeable. There is a particular reaction called an electron-positron annihilation. The positron is the electron's antiparticle, you may or may not have heard of antimatter, it has the same mass but an opposite charge, 1+. When a normal matter particle, the electron, meets with its antiparticle, the positron, they completely annihilate each other. What happens is the particles become energy, all of there mass becomes energy, in the form of photons.

    Another situation is called pair production, its when a high energy photon converts into an electron-positron pair. In that situation energy becomes matter!

    Each particle contains an amount of energy depending on its mass, the speed of light is large number, 299,792,458 m/s, squared makes it a much larger number. So when you calculate the amount of energy a certain chunk of matter contains with E=mc2, you'll find that a small chunk of matter equals an extremely large chunk of energy!

    Sadly, we have no practicle way to convert matter into energy, if we could we would have a practically endless supply of energy. To give you an idea of how much energy can be created in this fashion I'll give you a fun fact, the Sun's energy comes from conversion of matter into energy, around 4 million tonnes of matter becomes energy every second in the Sun. On Earth we have turned small portions of an objects mass into energy, this is what occurs during a nuclear reaction, i.e. nuclear weapons, the biggest explosives on Earth.
  6. Feb 5, 2009 #5
    Thanks very much for the response sir, that was very informative, so I guess what your saying is that the E=mc^2 equation is not really an equation as far as the mathematical definition goes and its really like more of a metaphor written as an equation?

    Like I have an equation X + Z = A, and I define the variables by saying X equales 1 apple and Z equales 2 apples and A equales the sum of X+Z. Then I can just insert the quantitive value of each defined variable and say that 1 + 2 = 3 and therefore the equation means 1 apple plus 2 apples equals 3 apples.

    Is E=mc^2 not really an equation in this sense and its just more of an abstract metaphor made to look like a math equation to express that energy and mass are the same thing?

    I can easily insert the known speed of light in this equation to take place of the “C” variable but…

    Is there not a real quantitive value you can insert for the “E” like volts or coulombs or electrons or something?

    Is there not a quantitive value you can insert for “M” like pounds ounces feet or miles?

    I have taken a lot of math but never saw an equation you cant do this with, how can you even use this equation to solve any kind of math problem if you cant assign quantitive values to the variables, its really just a fake equation then right? LOL
  7. Feb 6, 2009 #6


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    Staff: Mentor

    You have to use consistent units. We usually use a system of metric units.

    For example, (1 kilogram) times (300,000,000 meters per second) squared = 9,000,000,000,000,000 joules.

    Or (1 gram) times (30,000,000,000 centimeters per second) squared = 900,000,000,000,000,000,000 ergs.
  8. Feb 6, 2009 #7
    It's a real equation, for example, when the atomic bomb exploded, some amount of mass m "disppeared" because it was converted into energy E.

    If you put together a nucleus of an atom from the parts that it's made of, the nucleus will have less mass than the total of the masses of the parts that it was made of. Where did the missing mass go? It became the binding energy that holds the nucleus together.
    Last edited: Feb 6, 2009
  9. Feb 6, 2009 #8
    Your right, you have to use consistent units…

    Energy mass and speed are not even consistent units to start with that’s the problem, that’s like trying to add up 5 miles an hour plus 5 pounds of weight for a sum of 10 volts.
    Last edited: Feb 6, 2009
  10. Feb 6, 2009 #9
    Yeah all that might be true but true or not, the existence of the facts you have stated here dose not demonstrate the existence of E=mc^2 as a real equation…
  11. Feb 6, 2009 #10


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    Actually, the units do hold. 1 joule is 1 kilogram*(meter2)*(second-2). This is true with several other energy equations, if you bother to work them out. For example:
    Kinetic energy:

    Potential energy:

    Do you question the validity of either of these? If not, why question the validity of E=MC2?

    Incidentally, if you work that out into other units, 1 kilogram*meter2*second-2 is also equal to 1 newton*meter (because a newton is a kilogram*meter*second-2). This may make this unit easier to visualize - one joule of energy is the equivalent of an application of 1 newton of force over a distance of 1 meter. This may help show how the units actually work out.
  12. Feb 6, 2009 #11
    If you want an example: calculate the energy that will be released if four hydrogen atoms undergo fusion into a helium atom. Solution:

    E = [(4)(mass of one hydrogen nucleus) - (mass of one helium nucleus)] c^2

    ... where reference tables contain the values to be substituted.

    If you're looking for a proof, I don't know about that. I'm just accustomed to solving problems and later being told that I have obtained the right answers.
  13. Feb 6, 2009 #12
    1+1 dose not equal 2 because 3+7 equals 10.

    I can explain why 1+1=2 without referring to any other equations… 1+1 = 2 because when you add 1 to 1 you get a total of 2.

    Please don’t make my point more complicated by bringing in more equations that are fake for the same reason…




    These units of measure do hold up in this equation then explain how by telling me exactly how you can add, multiply, or divide a unit of mass with a unit of speed and come up with a value in energy by visualizing reality and not using an equation.

    Maybe you can explain how I could multiply 5 apples by 5 pounds of apples and come up with a group of apples moving at 25 miles an hour.

    Sense you say all these equations do hold then you should be able to explain how to add, subtract, multiply, or divide weight, length, and velocity with one another.


    What is 1 pound minus 1 mile?
    What is 1 pound times 1 mile?
    What is 1 pound times 1 mile an hour?

    For any of these equations to be real you have to explain how to do this by visualizing reality and not using an equation. Example : An apple and another apple together means both apples. (1+1=2)
    Last edited: Feb 6, 2009
  14. Feb 6, 2009 #13


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    You cannot add or subtract with the wrong units. However, your other two examples are actually doable. The second one, asking what is a pound times a mile, is quite simple. It is a unit of energy, assuming the pound is used as a force (rather than a mass). This is for the same reason as explained above - if you apply a set force over a set distance, you are doing work. This work is directly proportional to the change in kinetic energy. In this case, 1 pound force* 1 mile is roughly equal to 7200 joules.

    The third one, asking what is one pound time 1 mile per hour, actually becomes a power. The 1 pound* 1 mile portion is an energy, as explained above, and the per hour specifies that it is actually an energy per unit time, or a power. Specifically, it comes to about 2 watts.

    Now, the reason you can do this is because it is multiplication and division. With addition and subtraction, the units must match. However, with multiplication and division, they don't need to match, as you also multiply the units. For example, 1 newton* 1 meter is equal to (1*1) (newton*meters). For convenience, we have given this unit of the newton*meter a name (the joule), but it is still at its root a derived unit, that has several other units in the definition.

    Perhaps the easiest way of seeing this is with a force - if you take a mass and you multiply it by an acceleration, you get a force as the result. The reason for this is that by doing this multiplication, you are in essence saying that you have a mass that is accelerating. By F=MA, you have all of the information required to fully solve for the force. In the case of standard, SI units, 1 kg* 1 meter per second per second is equal to 1 newton. Therefore, a 1 newton force will accelerate 1 kilogram at a rate of 1 meter per second per second.
  15. Feb 6, 2009 #14
    Hey thanks for the informative answer, im trying to follow this…

    Check this out…

    One mile is 1609.344 meters. (lets just say 1609 meters )
    There are 453.59237 grams in one pound.
    What % of a pound is 102g? about 22.48% ( Lets say about 0.20 of a pound. )

    Ok lets say a joule is about 102g or about 0.20 pounds of force pushing for one meter strait up against earths gravity at sea level…

    So if the equation is

    1 pound of force * 1 mile = 7200 joules

    What two numbers are you multiplying to get 7200?
  16. Feb 6, 2009 #15


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    1 joule is 1 newton * 1 meter

    Therefore, to get joules from pounds of force and miles, you need to convert them to standard SI first:

    1 pound force = 4.45 newtons
    1 mile = 1610 meters

    So, 4.45 newtons * 1610 meters = 7164 newtons*meters = 7164 joules (I rounded to 2 significant figures in the previous post). Typically, for an answer in standard SI units, you need to use all of the input values in SI as well, hence newtons and meters (though the newton is a derived value equal to 1 kilogram * 1 meter per second squared)

    Oh, and the direction of the force is irrelevant - 1 joule is 1 newton of force acting over 1 meter, whether it is lifting something against earth's gravity or whether it is instead pushing it sideways across the floor.
  17. Feb 7, 2009 #16
    N is the force of Earth's gravity on an object with a mass of about 102 g (1⁄9.8 kg) (such as a small apple).

    Joule is The work done by a force the of Earth's gravity on an object with a mass of about 102 g (1⁄9.8 kg)(1 N) traveling through a distance of one meter

    1 joule = 1 newton * 1 meter

    102 g of force traveling through one meter = 102 g of force times 1 meter.

    102 g of force times 1 meter = 102 g of force traveling through one meter

    To say that 102 grams of force times 1 meter equales something, then 102g plus 1 meter must equal something also.

    Because 1 times 1 equals 1 and 1 plues 1 equals 2. If you can multiply 2 values together and get an answer you have to be able to add the same values to get an answer
    or they are not real values to begin with.

    1 newton * 1 meter = Joule

    102 g of force times one meter = Joule

    102 g of force times 1 meter = a fake value that cannot have phisical existance. Its derived from something that exists and something that dose not exist at all.

    You cant realy even say “102 g of force times 1 meter equals (anything)” because your trying to apply a mathimatical function between a value of moving mass that exists and a value of distance where distance dose not exist at all. Its impossible to multiply mass times speed without first inventing a fake value dirived from something that exists and something that dose not exist at all. If this could be done in reality then you would be able to add 102g of force to 1 meter…but you cant because the meter dose not exist phisicaly like the force.

    So I guess E=mc^2 is using these fake joules to say you can multiply mass with speed, or something, what ever it means you should be able to express it in plain english without using an equation of fake values.

    I can see how joules can be usful for engenering but…

    I think that you should be able to explain E=mc^2 in a new equation that dose not use fake values or use incompatible units together.
  18. Feb 7, 2009 #17
    If Joules looks like a fake unit to you, perhaps that's because there are always some arbitrary choices to be made in setting up any system of units. The kilogram is defined arbitrarily as the mass of a certain cylinder of metal that's stored in a vault. That's just a policy choice, not a scientific necessity.

    I have seen numerous systems of units used with E=mc^2.

    Before the meter-kilogram-second system of units was adopted by the "international system" conference, it was more customary to calculate e=mc^2 with e measured in ergs, m in grams, and c in cm/s.

    When working with atoms they will often use mass measured in atomic mass units (amu), and energy E measured in megaelectronvolts (MeV), and simply treat c^2 as a conversion factor by writing this in the reference tables: c^2 = 931.5 MeV/amu.
  19. Feb 7, 2009 #18


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    Correct so far, though a joule is the work done by any force of 1 newton acting over 1 meter. It isn't specific to gravity (though that example you used is a good way of visualizing the amount of force and energy involved).

    Sort of, though you really shouldn't think of grams (which are mass) as a force - use Newtons.

    Not necessarily.

    This doesn't necessarily follow. Just because you can both add and multiply simple numbers doesn't mean that anything that you can multiply can also be added. For example, it is fairly easy to see that if you take some distance, and you divide it by a time, you will get a speed. If you travel 10 meters in 2 seconds, you have traveled at an average speed of 10 meters/2 seconds = 5 meters/second. This does not mean that you are able to add 10 meters plus 2 seconds though. What is the sum of 10 meters and 2 seconds? The question is somewhat meaningless, as the units differ. Other units do not work out as intuitively as this, but they do still actually work.

    As I explained above, it is not a fake value, but a real value, just as real as speed (and you should really stop thinking of mass as force). Not only is it a real value in fact, it is a useful value, as it allows for simple calculations that show what will happen in various simple dynamic situations, such as the interaction between a mass and a spring.

    Actually, I can say that a force of 1 newton times 1 meter is equal to 1 joule. I am applying a mathematical function to a relation that it is relevant to. You need to stop thinking that just because some operations are possible that all are - in most cases, the rules for multiplication and division are the same, as is true for addition and subtraction. Between multiplication and addition however, the same rules do not always have to apply, because they are different operations. As for the meter not existing, are you saying that size is imaginary? I'm not quite sure where you're trying to come from here...

    Actually, E=mc2 is describing the actual amount of energy that could be released if mass were completely converted to energy (Note: it doesn't need to be in joules, however, if you use standard SI units, that is what E comes out in. If you use different units for M and C, E will be in different units as well). The units that it uses are the same as used in formulas for energy in basically all of classical physics, and can easily be shown to work.

    Have you ever taken a course on algebra? A good analogy to the units as far as rules for use are concerned is the set of rules for variables in algebra. For example, if you have 5x and multiply it by 3y, you get (5*3)(x*y), or 15xy. Even though these have different variables, they can be multiplied by or divided by with ease, and the math still works out. However, you cannot add 5x and 3y - the simplest form you can reduce it to is simply the statement 5x+3y - it does NOT turn into (5+3) + (x+y), (5+3)(x+y), (5+3)(x*y) or any other combination. This is the same as the rules for different units - you can multiply a meter and a kilogram together to get 1 kilogram*meter. However, you cannot simply add 1 kilogram and 1 meter.
  20. Feb 8, 2009 #19
    Your just converting data into meters per second – What you fail to see is your dividing 10 meters by 2 and getting 5 meters per second, your not dividing 10 meters by 2 seconds and getting 5 meters per second.

    10 meters divided by 2.

    Is not the same as

    10 meters divided by 2 seconds.

    10 meters divided by 2 = 5 meters and 10 meters divided by 2 seconds = Nothing, because its not possible.

    just like its impossible to add

    10 meters plus 2 seconds = nothing.

    You cant use incompatible units without getting a fake answer unless you invent fake values like joules.
  21. Feb 8, 2009 #20
    Your applying a mathematical function to a relation relevent to fake numbers.

    You cant compare physical mass that exists with units of measure that don’t exist and use them as if they are compatible.

    You can add length that dose not physically exist like mass with itself ( 1 mile plus 1 mile ) or you can add mass that dose exist with itself. ( 1 apple plus 1 apple ) You cant add/subtract/multiply/divide something that exists with something that doesn’t without inventing fake values that are the product of something that exists and something that dose not, thats impossible.
  22. Feb 8, 2009 #21


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    Correct. 10 meters divided by 2 is not the same as 10 meters divided by 2 seconds. 10 meters divided by 2 is 5 meters. 10 meters divided by 2 seconds is 10 meters per second.

    Actually, see above.

    I will repeat myself. Have you taken an algebra course?
  23. Feb 8, 2009 #22


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    Staff: Mentor

    This is covered in every introductory physics textbook on the planet. This is not a forum for arguing seriously against basic stuff like this.
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